Hi all,
Im working with FPU.
I have a real10 number for exemple Num1=4.0000000 in 80 bit format
This number is 4 but in real10 format.
If i truncate it to an integer number i get Num2=4 in dword format.
;
So i have one real10 number Num1=4.0000000 in 80 bit format (it is 4)
and an integer number Num2=4 in dword format
;.........
Num1 dt ???
Num2 dd ???
;.........
Now i want to compare Num1 with Num2 but i never get Num1 equal Num2
and they are equal ! They are the same number: 4 !
How to compare it corretly to get Num1 = Num2 ?
Thanks
Iam doing something like this:
mov eax, offset Num1
fld tbyte ptr [eax]
mov eax, offset Num2
fild dword ptr [eax]
...
fcom
...
hi Rui
you are trying to compare apples with oranges :P
probably the best way is to convert the dword integer(s) to real10, then compare
the case for the value 4 is somewhat special because it is a power of 2, however
if you do the same thing for the value 5.0, the results will be different
that is because the value 5.0 in real10 form is not exactly 5.0 - it might be something like 4.999999999999999999
Hi dave
Ok ! :P
«probably the best way is to convert the dword integer(s) to real10»
Ok i have Num2=4 in dword format. Tell me what to do to convert it
to to real10. How to code it ?
thanks
hi RuiLoureiro,
your above code looks right. Probably you miss the point, that fcom does not set the flags.
It is recommended to use fcomi[p], which does this.
;compare a,b
f[i]ld b
f[i]ld a
fcomip st,st(1)
fstp st
;use je,ja,jb,jea,...
use FILD to load the integer into an FPU register
if you haven't already done so, i suggest you look over Ray's FPU tutorial :U
http://www.ray.masmcode.com/tutorial/index.html
fild Num2
once it is loaded into the FPU, you can use one of the compare instructions (chapter 7)
to set up the FPU for 80-bit precision, you can just use FINIT
this also ensures that the 8 FPU registers are empty and available for use
there have been a few recent discussions on this subject :bg
you should be able to find all kinds of code, ideas, and theories - lol
http://www.masm32.com/board/index.php?topic=12998.0
http://www.masm32.com/board/index.php?topic=18253.0
I get equal for 4 or 5 (and for every other pair that I tried).
;==============================================================================
include \masm32\include\masm32rt.inc
.686
;==============================================================================
SHOWFLAGS MACRO
push ebx
pushfd
pop ebx
;------------------------------------
; This from Raymond's fputute:
; ZF PF CF
; If ST(0) > ST(i) 0 0 0
; If ST(0) < ST(i) 0 0 1
; If ST(0) = ST(i) 1 0 0
; If ST(0) ? ST(i) 1 1 1
;------------------------------------
.IF ebx & 64
printf("1 ")
.ELSE
printf("0 ")
.ENDIF
.IF ebx & 4
printf("1 ")
.ELSE
printf("0 ")
.ENDIF
.IF ebx & 1
printf("1\n\n")
.ELSE
printf("0\n\n")
.ENDIF
pop ebx
ENDM
;==============================================================================
.data
n1 REAL10 4.0
n2 dd 4
n3 REAL10 5.0
n4 dd 5
.code
;==============================================================================
start:
;==============================================================================
fld n1
fild n2
fcomip st(0),st(1)
fstp st(0)
SHOWFLAGS
fld n3
fild n4
fcomip st(0),st(1)
fstp st(0)
SHOWFLAGS
inkey
exit
;==============================================================================
end start
yah - the assembler assigns the same value to the float as the FPU does with FILD :bg
but - if it is the result of some previous calculation, things can get a little difficult
comparing floats is no simple subject - refer to the threads i posted above
qWord,
Thanks
I get this at instuction fcomip st,st(1) :
Microsoft (R) Macro Assembler Version 6.14.8444
Copyright (C) Microsoft Corp 1981-1997. All rights reserved.
Assembling: ... : error A2085: instruction or register
not accepted in current CPU mode
_
Assembly Error
Prima qualquer tecla para continuar . . .
What to say ?
see MichaelW' code: .686 is required
Run it with Olly, and watch what the FPU does. Or use MasmBasic's Fcmp macro.
There is more to be tested here than just the FPU comparisons. One possibility for the values not comparing as equal would be if MASM were not encoding the FP values correctly.
i returned to my work just now
qWord,
Thanks it works
JJ2007,
I dont want to test it
MichaelW,
Thanks
«There is more to be tested here than just the FPU comparisons.
One possibility for the values not comparing as equal would be if MASM
were not encoding the FP values correctly.»
This is not a MASM problem. :green2
my problem:
I type some numbers (2 and (2 and 2)). Each 2 is coded
in real10. And we calculate 2^2 as a real10 number.
I get 4 in real10. Nothing wrong.
Now i need to calculate 2^4. 2 is real10, 4 is real10
But i want to calculate this 2^4 like 2 in real10 and
i want to get 4 in integer32 to use another formula.
i think i could truncate the real10 4 to integer32
but only if the real10 4 is equal to integer32 4.
I get integer32 4 is greater than real10 4.
May be i cannot do this this way !
Any help ?
;------------------------------
I have one formula to compute X^Y if X and Y are both real10
and i have another formula to compute X^Y if X is real10
and Y is integer32.
But i dont want to code Y in integer32, this is the problem
Did you understand the problem ?
;------------------------------
Quote from: RuiLoureiro on March 17, 2012, 06:45:56 PM
I dont want to test it
You don't have to. But you will never understand what your code does if you don't learn at least to use Olly.
Quote from: RuiLoureiro on March 17, 2012, 06:45:56 PM
Now i need to calculate 2^4. 2 is real10, 4 is real10
But i want to calculate this 2^4 like 2 in real10 and
i want to get 4 in integer32 to use another formula.
i think i could truncate the real10 4 to integer32
but only if the real10 4 is equal to integer32 4.
I get integer32 4 is greater than real10 4.
Do us a favour and read Ray's tutes.
well - you may be able to do it entirely with integers :P
how large can the numbers be ?
http://www.masm32.com/board/index.php?topic=12363.msg100575#msg100575
JJ,
i read the tutes and i didnt find any help to the :green2
problem
Dave,
they are as large as real10 numbers
oh - ok
well - the routine i linked will generate integer exponentials to full precision
RuiLoureiro,
you want to test whether the content of a FPU register is an integer value or not?
Dave,
Ok but it doesnt help me
qWord,
you want to test whether the content of a FPU register is an integer value or not?
no.
Rui,
Post:
1. full code
2. your expected value
3. what the FPU produces against your expectations.
Everything else is a waste of time - your time and our time.
:thumbu
jj,
1. thank you for your answer :thumbu
2. Sorry, full code is too long !
3. I have one formula to compute X^Y if X and Y are both real10
and i have
another formula to compute X^Y if X is real10 and Y is integer32.
But i dont want to code Y in integer32,
this seems to be the problem. But i dont know if i must code it
in integer32.
I know how to code Y in integer32 but i dont want to do that
if i can solve the problem in a different way.
If Y is in Real10, i am thinking that when i truncate Y to
integer32 i get Yi (Yi is an integer32).
Yi = Trunc (Y)
Now:
mov eax, offset Y ; Y is 4 in real10
mov ebx, offset Yi ; Yi is 4 in integer32
fld tbyte ptr [eax]
fild dword ptr [ebx]
fcomip st,st(1)
jne _isnotequal
je _isequal ; why not ?
It seems to be clear, no ? I get _isnotequal.
Did you understand the problem ?
4. The question is: Can i solve the problem this way
or i need to code Y in integer32 to use the 2nd formula ?
Thanks
... so you want to determinate if y is an integer or not and then select the corresponding algorithm ::)
Quote from: qWord on March 17, 2012, 09:31:30 PM
... so you want to determinate if y is an integer or not and then select the corresponding algorithm ::)
Yes
1) use FINIT once at the beginning of the program - it sets up the FPU for 80-bit precision
when you start a program under windows, it is set up for 64-bit precision
2) use REAL10 instead of DT or TBYTE PTR
3) this code leaves one of the FPU registers loaded - make sure you FFREE it :P
.XCREF
.NOLIST
INCLUDE \masm32\include\masm32rt.inc
.686
.LIST
;#################################################################################
.DATA
Y real10 4.0
Yi dd 4
;*********************************************************************************
; .DATA?
;#################################################################################
.CODE
;*********************************************************************************
_main PROC
finit
fld real10 ptr Y
fild dword ptr Yi
fcomip st,st(1)
jz equal
not_equal:
print chr$('Not Equal'),13,10
jmp short exit00
equal:
print chr$('Equal'),13,10
exit00:
inkey
exit
_main ENDP
;#################################################################################
END _main
result:
Equal
Press any key to continue ...
Quote from: RuiLoureiro on March 17, 2012, 09:32:28 PM
Quote from: qWord on March 17, 2012, 09:31:30 PM
... so you want to determinate if y is an integer or not and then select the corresponding algorithm ::)
Yes
see what I've found in my FPU box:
fld ...
sub esp,12
fst[p] REAl10 ptr [esp]
movzx eax,WORD ptr [esp+8]
and eax,7FFFh
mov ecx,eax
xor eax,7FFFh
.if !ZERO? ; number is not NaN,INFINTE,...
mov eax,DWORD ptr [esp]
or eax,DWORD ptr [esp+4]
or eax,ecx
setz al
.if !ZERO? ; value != +-0.0
xor eax,eax
sub ecx,3fffh
.if !CARRY? ; negativ exponent => not an integer
test ecx,0ffffffc0h
setnz al
.if ZERO? ; (exponent >= 64) ? integer : continue check
; clear out all integer bits and test for zero.
; (could be replaced by common x86 code...)
lea ecx,[ecx+1]
movq xmm0,QWORD ptr [esp]
movd xmm1,ecx
pcmpeqb xmm2,xmm2
pxor xmm3,xmm3
psrlq xmm2,xmm1
pand xmm0,xmm2
pcmpeqd xmm0,xmm3
pmovmskb eax,xmm0
movzx eax,al
xor eax,0ffh
setz al
.endif
.endif
.endif
.endif
add esp,12
; eax = 0/1
; eax=0: floating point value , eax != 0: integer I've not used this offten, but AFAIKS it works well.
The problem with fcomip is that the two numbers need to be really, really equal. And not every integer can be represented exactly in REAL10 format. To illustrate this, here an example where the 19th digit is a tick off:
include \masm32\MasmBasic\MasmBasic.inc ; download (http://www.masm32.com/board/index.php?topic=12460)
.data
Y REAL10 4.000000000000000001
Yi DWORD 4
Init
mov eax, offset Y ; Y is 4 in real10
mov ebx, offset Yi ; Yi is 4 in integer32
fld REAL10 ptr [eax]
fild dword ptr [ebx]
deb 4, "FPU", ST(0), ST(1)
Print "fcomip, REAL10: ", Tb$
fcomip st,st(1)
.if Zero?
PrintLine "equal"
.else
PrintLine "not equal"
.endif
mov ecx, Yi
Print "Fcmp, top precision:", Tb$
Fcmp Y, ecx, top
.if Zero?
PrintLine "equal"
.else
PrintLine "not equal"
.endif
Print "Fcmp, high precision:", Tb$
Fcmp Y, ecx, high
.if Zero?
PrintLine "equal"
.else
PrintLine "not equal"
.endif
Inkey "ok"
Exit
end start
FPU
ST(0) 4.00000000000000000
ST(1) 4.00000000000000000
fcomip, REAL10: not equal
Fcmp, top precision: not equal
Fcmp, high precision: equal
The debug macro "sees" the same number, but fcomip and the Fcmp macro in top precision mode see it as different. Fcmp in "high" precision, i.e. roughly REAL8, treats them as equal.
Now the obvious question is "what for"? Which precision does your app need? The "not equal" of fcomip would be misleading in most instances.
The question would be irrelevant if 4.0 came directly from memory - you declare Real10 4.0, and it works. However, if 4.0 are the result of calculations, then they are most likely a tick off, and you will almost never see "equal". Especially if the app does graphics, with a resolution of e.g. 2000 pixels, then you would need a much lower precision in order to see an equal where you want to see an equal.
P.S.: When testing this, I noticed a little bug in Fcmp - it would load a DWORD global variable as REAL4. The bug is fixed (http://www.masm32.com/board/index.php?topic=12460.msg156620#msg156620), and the example above works also with the old version because I used mov ecx, Yi to circumvent this problem.
This precision problem can be solved by using a threshold-factor:
pseudo code:
if abs(a-b) <= abs(a*factor) then equal(requires: a != 0)
The SmplMath (http://sourceforge.net/projects/smplmath/)-macro faEQ(a,b,f) supports this method:
Quoteinclude \masm32\include\masm32rt.inc
.686
.mmx
.xmm
include \macros\smplmath\math.inc
.code
main proc
LOCAL fSlvTLS()
print "exact compare: ",13,10
.if fEQ(1.01,1.02)
print " fEQ( 1.01, 1.02): equal (exact)",13,10
.else
print " fEQ( 1.01, 1.02): not equal (exact)",13,10
.endif
print chr$(13,10)
print "compare using a threshold factor: ",13,10
.if faEQ(1.01,1.02)
print " faEQ( 1.01, 1.02, 1.0E-2): equal (+-1%) ",13,10
.else
print " faEQ( 1.01, 1.02, 1.0E-2): not equal (+-1%)",13,10
.endif
.if faEQ(1.01,1.02,1.0E-3)
print " faEQ( 1.01, 1.02, 1.0E-3): equal (+-1 per mille) ",13,10
.else
print " faEQ( 1.01, 1.02, 1.0E-3): not equal (+-1 per mille)",13,10
.endif
print chr$(13,10)
inkey
exit
main endp
end main
exact compare:
fEQ( 1.01, 1.02): not equal (exact)
compare using a threshold factor:
faEQ( 1.01, 1.02, 1.0E-2): equal (+-1%)
faEQ( 1.01, 1.02, 1.0E-3): not equal (+-1 per mille
Quoteanother formula to compute X^Y if X is real10 and Y is integer32.
I will have to assume at this point that what you mean by
integer32 is a 32-bit integer, i.e. a dword.
ALL
dwords are represented
exactly in REAL8 and REAL10 format.
In REAL8 format, the least significant 20 bits would ALWAYS be 0, while in the REAL10 format, the least significant 32 bits would ALWAYS be 0.
One way to check if a REAL10 number in memory is effectively a 32-bit signed integer could be done as follows:
.data
myREAL10 REAL10 ...
.code
mov eax,dword ptr myREAL10 ;get the lower 32 bits
test eax,eax
jnz not_a_dword_integer
mov eax,dword ptr myREAL10[32] ;get the first 32 bits of the significand field
bsr ecx,eax
mov eax,31
sub eax,ecx ;number of active bits
add eax,3FFFh ;minimum biased exponent for those bits to be an integer
mov edx,dword ptr myREAL10[48] ;get the upper 32 bits
shl edx,1 ;get rid of the sign bit
shr edx,17 ;retain only the biased exponent
cmp edx,4020h
jae not_a_dword_integer ;the value of myREAL10 would be larger than 2^32
cmp edx,eax
jb not_a_dword_integer
;is a dword integer
...
not_a_dword_integer:
...
Quote 3. I have one formula to compute X^Y if X and Y are both real10
and i have
another formula to compute X^Y if X is real10 and Y is integer32.
I don't know what "formulas" you are using to compute the powers of REAL numbers but, in my opinion, the simplest way is to use the FPU where it doesn't matter at all whether the "Y" is a float or an integer. Maybe you can enlighten us on those different formulas.
Maybe you can compare the real10 numbers outside the FPU...
Num1 XOR Num2
Should give you and idea if it's at least equal (zero result).. if not then maybe it's the compiler/assembler not assigning equal values
:8)
SmplMath and MasmBasic use two different methods to accommodate rounding errors:
.if faEQ(100.01,100.02,1.0E-3) ; +- 1.0E-3
print "is equal"
.endif
Fcmp FP8(100.01), FP8(100.02), 6 ; ca six digits
.if Zero?
print "is equal"
.endif
In other words, SmplMath sets an absolute error range, while MasmBasic uses a relative precision. What is "better" depends on your app.
In both cases, the point here is that the FPU, when reporting "equal", means "equal relative to my internal 64-bit precision".
The problem is you won't notice the difference. Or, more precisely: most of the time, you won't notice... :bg
test4=12345678 ; try taking away the 8...
include \masm32\MasmBasic\MasmBasic.inc ; download (http://www.masm32.com/board/index.php?topic=12460)
include \macros\smplmath\math.inc
.data?
MyReal10 REAL10 ?
.code
main proc
LOCAL fSlvTLS() ; otherwise smplmath says it's not thread-safe
print "exact compare: ",13,10
.if fEQ(1.01,1.02)
print " fEQ( 1.01, 1.02): equal (exact)",13,10
.else
print " fEQ( 1.01, 1.02): not equal (exact)",13,10
.endif
print chr$(13,10)
print "compare using a threshold factor: ",13,10
.if faEQ(1.01,1.02)
print " faEQ( 1.01, 1.02, 1.0E-2): equal (+-1%) ",13,10
.else
print " faEQ( 1.01, 1.02, 1.0E-2): not equal (+-1%)",13,10
.endif
.if faEQ(1.01,1.02,1.0E-3)
print " faEQ( 1.01, 1.02, 1.0E-3): equal (+-1 per mille) ",13,10
.else
print " faEQ( 1.01, 1.02, 1.0E-3): not equal (+-1 per mille)",13,10
.endif
.if faEQ(100.01,100.02,1.0E-3)
print " faEQ( 100.01, 100.02, 1.0E-3): equal (+-1 per mille) ",13,10
.else
print " faEQ( 100.01, 100.02, 1.0E-3): not equal (+-1 per mille)",13,10
.endif
print chr$(13,10), "Same with MasmBasic Fcmp:", 13, 10
Fcmp FP8(100.01), FP8(100.02), high ; ca REAL8
.if Zero?
print " Fcmp 100.01, 100.02, high: equal ",13,10
.else
print " Fcmp 100.01, 100.02, high: not equal ",13,10
.endif
Fcmp FP8(100.01), FP8(100.02), 5 ; ca. five digits
.if Zero?
print " Fcmp 100.01, 100.02, 5: equal ",13,10
.else
print " Fcmp 100.01, 100.02, 5: not equal ",13,10
.endif
Fcmp FP8(100.01), FP8(100.02), 6 ; ca six digits
.if Zero?
print " Fcmp 100.01, 100.02, 6: equal ",13,10
.else
print " Fcmp 100.01, 100.02, 6: not equal ",13,10
.endif
Fcmp FP8(100.01), FP8(100.02), low ; ca REAL4
.if Zero?
print " Fcmp 100.01, 100.02, low: equal ",13,10, 10
.else
print " Fcmp 100.01, 100.02, low: not equal ",13,10, 10
.endif
test4=12345678 ; try taking away the 8...
print "little test x=sqrt(x/7)^2*28/x (should be exactly 4):", 13, 10, "for x="
print str$(test4), ":", 13, 10
void Str$(0) ; initialise some MB routines
push test4
push 7
push 28
push test4
fild stack
pop eax
fild stack
pop eax
fild stack
pop eax
fild stack
pop eax
deb 4, "FPU", ST(0), ST(1), ST(2), ST(3) ;
fdivr ; 4/7=0.57
fsqrt ; 0.7
fld st
fmul ; 4/7
fmul ; 4/7*28=16
fdivr ; 16/4=4
fstp MyReal10
deb 4, "R10", MyReal10
Fcmp MyReal10, 4, top
.if Zero?
print "4 is 4", 13, 10
.else
print "4 is not 4", 13, 10
.endif
inkey chr$(13, 10, 10)
exit
main endp
end main
Output:
exact compare:
fEQ( 1.01, 1.02): not equal (exact)
compare using a threshold factor:
faEQ( 1.01, 1.02, 1.0E-2): equal (+-1%)
faEQ( 1.01, 1.02, 1.0E-3): not equal (+-1 per mille)
faEQ( 100.01, 100.02, 1.0E-3): equal (+-1 per mille)
Same with MasmBasic Fcmp:
Fcmp 100.01, 100.02, high: not equal
Fcmp 100.01, 100.02, 5: equal
Fcmp 100.01, 100.02, 6: not equal
Fcmp 100.01, 100.02, low: equal
little test x=sqrt(x/7)^2*28/x (should be exactly 4):
for x=12345678:
FPU
ST(0) 12345678.0000000000
ST(1) 7.00000000000000000
ST(2) 28.0000000000000000
ST(3) 12345678.0000000000
R10 MyReal10 4.00000000000000000
4 is 4
Thanks for your replies
I need to give answers to all of you. Wait please
Dave,
1) use FINIT once at the beginning of the program - it sets up the FPU for 80-bit precision
when you start a program under windows, it is set up for 64-bit precision
2) use REAL10 instead of DT or TBYTE PTR
3) this code leaves one of the FPU registers loaded - make sure you FFREE it
1. The first instruction is exactly FINIT
3. Every succesful calculation starts loading the FPU
with the first operand, next the second operand,
next the operation and end with store the result at a
REAL10 variable.
Example 1:
X dt 2
Y dt 5
Z dt ?
mov eax, offset X
mov ebx, offset Y
;
fld tbyte ptr [eax] ; st(1)
fld tbyte ptr [ebx] ; st(0)
fadd ; new st(0) = st(1)+st(0)
mov eax, offset Z
fstp tbyte ptr [eax] ; store st(0) and pop
What to FFREE ?
Example 2:
X dt 2
Y dt 5
Z dt ?
mov eax, offset X
;
fldpi ; st(2)
fld st(0) ; st(1) get a copy
fld tbyte ptr [eax] ; st(0)
fadd ; new st(0) = st(1)+st(0)
mov eax, offset Z
fstp tbyte ptr [eax] ; store st(0) and pop
fstp st ; remove source
What to FFREE ?
I think all this examples are correct and we dont need to free nothing,
we dont need to do finit againg. Correct Dave ?
qWord,
This precision problem can be solved by using a threshold-factor:
pseudo code:
if abs(a-b) <= abs(a*factor) then equal(requires: a != 0)
0. Thanks for your reply;
1. I dont know exactly what to do; What factor to use
2. I think you understood the problem;
I repeat it (this is exactly one example)
a) From keyboard we type a string "2^(2^2)";
b) The first thing to do is to solve (2^2);
c) We convert the first 2 to a real10 variable X
and the second 2 to a real10 variable Y;
d) We apply the first formula to get X^Y in a real10 variable Z;
So far so good!
Now i want to use the second formula X^Zi where Zi is an integer
( X is real10 and Zi dword )
or the first formula X^Z ( X is real10 and Z is real10 )
Well, if Z=X^Y is an integer than Z=Trunc(Z) or Z-Trunc(Z)=0
if Z=X^Y is not an integer than Z != Trunc(Z) or Z-Trunc(Z) != 0
Correct ?
I write 2 or 4 if integer and 2.0 or 4.0 if real, ok ?
Solving this by hand, we have 2.0^2.0 is a real 4.0.
So Z=4.0 and Trunc(4.0) = 4 than «Z is equal to Trunc(Z)»
FPU tell me «Z is lower than Trunc(Z)»
I dont know what is the result of ABS(Z-Trunc(Z)) using the FPU till now
Raymond,
Thank you for your replies
1. I know your fpulib: FpuAtoFL, FpuTrunc, FpuXexpY, FpuCos, etc. etc.
I read all what is written there.
2. I think that FpuXexpY solves X^Y if and only if X>0
it doesn't matter at all whether the "Y" is a float or an integer.
Am i correct ?
3. The problem is when X is negative. How do you solve it in that case ?
This answer your question, i think
jj,
Thanks for your reply
The problem is not with fcomip. If you read my answers maybe you
understand the problem
vanjast,
Thank you for your help
raymond,
Please, read the answer i gave to qWord.
Yes, integer32 is a dword.
I used the code you post and i got Z is not_a_dword_integer
It is 4.0... but not a_dword_integer
It doesnt solve my problem.
In your last post it seems you have only one formula to solve
the problem of (X)^(Y) where X and Y are integers or reals.
I am waiting your answer.
Thanks
Quote from: RuiLoureiro on March 18, 2012, 10:06:10 AM What to FFREE ?
the FPU has 8 registers
they behave differently from CPU general registers in many ways
one important difference is that each register is always marked as empty or full
if all 8 registers are full, an error (exception) will occur if you attempt to load another value
so - if you load (or push) 2 values, you should pop 2 to keep balance
this allows you to use the function repeatedly without filling all the registers
in the code i posted
finit
fld real10 ptr Y
fild dword ptr Yi
fcomip st,st(1)
jz equalFINIT empties all 8 registers
we then load 2 values with FLD and FILD
the FCOMIP instruction pops one of those values
but, we have left the other value in the FPU
if we want to use the code over and over without FINIT, we should FFREE st(0)
FINIT is a nice instruction :P
but, it takes a lot of clock cycles
Quote from: RuiLoureiro on March 18, 2012, 11:29:42 AM
The problem is not with fcomip.
if Z=X^Y is not an integer than Z != Trunc(Z)
The FPU will tell you Z != Trunc(Z) because of its precision limits. Your examples are confused, it would really help if you could isolate the problem and post the code. Can't be so difficult for a console app...
Quote from: jj2007 on March 18, 2012, 09:52:15 AM
In other words, SmplMath sets an absolute error range, while MasmBasic uses a relative precision.
no, SmplMath error range is also relative: e.g. if abs(a-b) <= abs(a*1.E-2) ...
This method fails, if a=0.
Interesting to see, that MB is analysing the fraction bits - vieleicht klau ich die Idee :bg
Quote from: qWord on March 18, 2012, 02:09:41 PMInteresting to see, that MB is analysing the fraction bits - vieleicht klau ich die Idee :bg
Finger weg! I just downloaded SmplMath, looks good :U I am in an advanced stage of rolling my own, so I will deliberately avoid looking at your sources. My tokeniser is working perfectly, but i got a little bit stuck with the preference order, and would need a) a full night of sleep, b) a full day of quiet coding, both of which won't happen anytime soon :(
Anyways, compliments for SmplMath :thumbu
Quote from: jj2007 on March 18, 2012, 02:16:00 PM
I am in an advanced stage of rolling my own, so I will deliberately avoid looking at your sources. My tokeniser is working perfectly, but i got a little bit stuck with the preference order, and would need a) a full night of sleep, b) a full day of quiet coding, both of which won't happen anytime soon :(
Anyways, compliments for SmplMath :thumbu
It will work out in the end - I needed 3 attempts and 5 years to get it.
:bg
Quote3. The problem is when X is negative. How do you solve it in that case ?
With X being negative, the result is a real number when Y is an integer. However, if Y is not an integer, the result is a complex number.
Can you handle complex numbers?
If yes, you could use the Zlib (similar to the Fpulib but for complex numbers) from http://www.ray.masmcode.com/complex.html
It would return the result with the correct sign for all cases, whether Y is integer or not.
Note: the downloadable file contains a dialog box which you can use to test any of the functions with any input parameters you want.
Here are a few examples of the previous post.
(-3.0)2 = 9 + 0(i)
(-3.0)2.2 = 9.070357505108625 + 6.590000471674018(i)
(3.0)2.2 = 11.21157845653966 + 0(i) = sqrt(9.0703575051086252 + 6.590000471674018(i)2)
(-3.0)2.5 = 0 + 15.58845726811990(i)
(3.0)2.5 = 15.58845726811990 + 0(i)
(-3.0)2.8 = -17.53465226998805 + 12.73967058793206(i)
(3.0)2.8 = 21.67402216752623 + 0(i) = sqrt((-17.53465226998805)2 + 12.73967058793206(i)2)
(-3.0)-2.8 = -0.037326574095097 + -0.027119343504831(i)
(3.0)-2.8 = 0.046138182948723 + 0(i) = sqrt((-0.037326574095097)2 + (-0.027119343504831(i))2)
raymond,
Quote
Can you handle complex numbers?
Yes, but in my present work i dont want to use it.
In any case, thank you for your help.
Quote
With X being negative, the result is a real number when Y is an integer
This is why we need 2 formulas:
1) (X)^(Y) X, Y real10
2) (X)^(Yi) X real10, Yi integer32
In general, X and Y are the results of expressions.
Because i am working with real10 numbers, i have a calculator
(a proc that does this) that gives me the result in real10.
For that reason, i want to get Yi from Y, making Yi=Trunc(Y)
and not from a particular procedure that calculate Yi directly.
The problem is to decide if Yi is the correct integer and if so
we use formula 2).
If Y is not an integer we use formula 1).
Have you any opinion about this ?
Quote
About FpuSin
Could you tell me why it gives Sin( pi ) = -0.00000... ?
(But FpuCos doesnt give it with pi/2)
And why not 0.00000... ?
Thanks
QuoteQuote
Can you handle complex numbers?
Yes, but in my present work i dont want to use it.
If Y is not an integer we use formula 1).
The only conclusion I can reach from the above quotes is that "formula 1" uses the absolute value of X and returns the length of the vector (without the angle) in the complex plane when X is negative. Otherwise, you don't have any choice but to use complex numbers to compute such powers and return both cartesian (or polar) coordinates in the complex plane.
raymond,
Quote
The only conclusion I can reach from the above quotes is that
"formula 1" uses the absolute value of X
1. Why do you say "uses the absolute value of X" ?
2. Could you give me an example where X is a negative real10 number
and Y is not integer and X^Y is a real10 number ?
3. What about FpuSin question ?
Quote from: RuiLoureiro on March 18, 2012, 09:59:18 PM 2. Could you give me an example where X is a negative real10 number
and Y is not integer and X^Y is a real10 number ?
e.g.: x
1/n ; n = +-1, +-3, +-5,... ; x can be any positive or negativ value
Quote from: RuiLoureiro
Could you tell me why it gives Sin( pi ) = -0.00000... ?
(But FpuCos doesnt give it with pi/2)
And why not 0.00000... ?
Unusually large, imprecise, value for PI?
22/7 springs to mind, but you should check the FPU constant for PI against your own.
Quote1. Why do you say "uses the absolute value of X" ?
Because the usual manner to raise a value X to a power Y when Y is not an integer is to compute the logarithm of X (in whichever base you want), multiply it by the power, and then get the antilog (in the same base) of the product. BUT, logarithms of negative numbers do not exist except with complex numbers (and you don't use them).
Quote2. Could you give me an example where X is a negative real10 number and Y is not integer and X^Y is a real10 number ?
Not without using complex numbers. But, you could express
both cartesian (or polar) coordinates of the result with REAL10 numbers. Only the powers of positive numbers can be expresed with a single REAL10 result (the coefficient of the imaginary component would always be 0 when computed as complex numbers).
Quote Could you tell me why it gives Sin( pi ) = -0.00000... ?
(But FpuCos doesnt give it with pi/2)
And why not 0.00000... ?
Regardless of how you assign the value of pi, it will always have some small error (either + or -). If the result of your computation with that pi value is so close to zero that it cannot be returned with the smallest value available according to the precision control of the FPU, it will be returned with +0 or -0 according to which side of the zero the result is closest (the FPU is designed to perform computations with a few extra bits over and above those you see in a REAL10).
BUT, not to worry, if you compare a -0 with a +0 with the FPU, they will be considered as EQUAL.
QuoteY is not integer and X^Y is a real10 number ?
Obviously, as the REAL10 exponent get closer and closer to an integer, the result will ressemble more and more the exponentiation by an integer. Here are a few examples to show how close it may become. You can draw your own conclusions.
X Y r i
-3.0 2.10 9.553465958876771 3.104109257329652
-3.0 2.01 9.094930212851599 0.285819696514787
-3.0 2.001 9.009848481854879 0.028305366921538
-3.0 2.0001 9.000988361193504 0.002827743984086
-3.0 2.00001 9.000098871207738 0.000282746445047
-3.0 2.0000000001 9.000000000988751 0.000000002827433
Hi all
I started writing something to solve a real expression a week ago.
What i want ? The answer is: i dont know exactly.
What to do ? The answer is: i dont know exactly.
What is the project ? The answer is: i dont know exactly.
Is it to solve all kind of expressions ? Probably not.
Is it to do the same thing as we know ? Probably not.
What is necessary ? I know: start it.
I decided to work with fpu and real10 numbers. Is it necessary to
work with complex numbers ? The answer is: i dont know exactly.
But i know one thing: i dont want to use it for now.
I have some experience with fpu, but not much. Well, i am going
to face some problems i know, but we learn with problems to solve.
So, i am in the stage where i go to define what and how to do.
qWord,
Yes, that is some particular cases to solve. Those and this
(X)^(m/n) m,n integer32
SQR(X)=sqr(X)=(X)^(1/2)
I am thinking to use one particular function to solve it.
That problem of Yi=Trunc(Y) is solved with a factor
The error Yi-Y is <1.0e-10 or less
I use this. After fcomip i remove source. Is this the best
place to remove it ?
; ---------------------------
; Move ErrorDelta to be st(1)
; ---------------------------
mov eax, offset ErrorDelta
fld tbyte ptr [eax]
mov eax, offset ExponentI
fild dword ptr [eax]
mov eax, offset ExponentZ
fld tbyte ptr [eax]
fsub
fabs
; -------
; Compare
; -------
fcomip st(0), st(1)
fstp st ; ««««« remove source
fwait
jpe short error
jb short _menor
ja short _maior
;je short _igual
;_igual:
_menor:
_maior:
Thanks
Clive and raymond,
This is the procedure.
_sin: fldpi
fadd st,st ;->2pi
fxch
@@: fprem ; reduce the angle
fsin
fstsw ax
fwait
shr al,1 ; invalid operation ?
jc _erro
sahf ;transfer to the CPU flags
jpe @B ;reduce angle again if necessary
fstp st(1) ;get rid of the 2pi
jmp _exit
...
_exit: mov eax, offset Z
fstp tbyte ptr[eax]
When we write sin(pi) the constant for PI
is loaded from fpu (fldpi) and stored at a varible X
fldpi
mov eax, offset X
fstp tbyte ptr[eax]
Before we go to _sin, we load it from X
Thanks :wink
raymond,
Quote
Because the usual manner to raise a value X to a power Y when Y is not an integer
is to compute the logarithm of X ...
Yes, it is what i know
Quote
BUT, logarithms of negative numbers do not exist except with complex numbers
Yes, it is what i know
And it tell us that we have other problems if X<0.
If X=0 we use the same procedure and fpu report error and than exit ...
Quote
You can draw your own conclusions.
I say the same you said: «as the REAL10 exponent get closer and closer
to an integer, the result will ressemble more and more the exponentiation
by an integer». So, make all sense to use the integer part of Y and
give a real10 result. It seems to me.
Thanks raymond :wink
Just to save you some precious time with unnecessary typing, assuming you are using MASM, you don't have to dereference variables to load the FPU if their size have already been declared in either global or local memory. For example,
mov eax, offset ExponentI
fild dword ptr [eax]
mov eax, offset ExponentZ
fld tbyte ptr [eax]
fsub
If ExponentZ was declared as a tbyte and ExponentI was declared as a dword, you could simply do:
fild ExponentI
fld ExponentZ
fsub
Even shorter:
fld ExponentZ
fisubr ExponentI
And it is a waste of your precious time to store the value of pi in memory. It is always available directly from the FPU whenever needed.
raymond,
Thank you for your reply
Quote
And it is a waste of your precious time to store the value of pi in memory.
It is always available directly from the FPU whenever needed.
It seems i need to store it to OperandZ. When we type an expression
it is saved in a string, and it can be "sin (2*pi*20/60)" or
"sin(pi)" or "sin(5.1*pi*20/70+...)". When we find "(" and ")"
a procedure should solve it and put the result in OperandZ. Then
a general procedure solves the function sin(OperandZ). I have
not a particular procedure to solve "sin(pi)" or other argument.
QuoteUsing the absolute value of X returns the length of the vector (without the angle) in the complex plane when X is negative.
For those interested in the subject, the angle in the complex plane for X^Y can easily be computed as follows when X is negative and Y is not an integer.
(In the complex plane, the angle is measured against the positive real axis and increases counterclockwise.)
- Compute the vector X^Y with the absolute value of X
- Get the nearest
even integer N less than Y
- Subtract it from Y
- The angle in radians = pi*(Y-N), and in degrees = 180*(Y-N)
For example, if Y = 3.12, the nearest even integer (less than Y) is 2
3.12 - 2 = 1.12
the angle = 1.12*pi = 3.5185837720 radians = 201.6 degrees
If Y = -3.12, the nearest even integer (less than Y) is -4
-3.12 - (-4) = 0.88
the angle = 0.88*pi = 2.7646015352 radians = 158.4 degrees
Hi all
This is the Calcula20.
Try to solve some expressions and say something
Did you find an error result ?
Please, tell me to help me
Thanks :U
Cool :U
I notice you return ERROR for X^Y when X is negative and Y is not an integer. :U :wink
I had a quick try with a relatively long expression (using *, /, -, sind, cosd, exp, sqr and a few pi) and it worked OK for it. :clap:
-------------------------------------------------------------------
Type an expression and press enter [ ex: -pi + 3.1*(7 - 2*8.3) ]
-3^2.5
This is your expression: -3^2.5
This is your Result: -15.5884572681
-3^2.5=-15.5884572681
---------------------------------------------
Type an expression and press enter [ ex: -pi + 3.1*(7 - 2*8.3) ]
also, how do you exit ? - i didn't see it :bg
very cool program :U
I tried a long expression for X^Y where X would be negative and Y not an integer and it returned ERROR.
Now, I tried Dave's attempt of -3^2.5 and it doesn't return ERROR but a "wrong" result.????????????
jj,
Thanks :wink
dave,
Thanks. :wink
It doesnt exit. Close the window !
raymond,
Thanks :wink
Quote
I tried a long expression for X^Y where X would be negative and Y not an integer and
it returned ERROR.
If X<0 and Y is not an integer it may be an ERROR or the prog
doesnt solve it correctly. Please tell me the case.
I didnt solve all cases. I have a lot of work to do in
the next days.
Quote
Now, I tried Dave's attempt of -3^2.5 and it doesn't return ERROR
but a "wrong" result.????????????
"wrong" result ? But as dave show, -3^2.5=-15.5884572681 is correct.
I dont know what case you are talking about.
QuoteBut as dave show, -3^2.5=-15.5884572681 is correct.
No it isn't.
-3^2.5 = -3^2 * -3^0.5
= 9 * ±1.7320508j
= ±15.58845jThere are 2 solutions, both are imaginary.
Paul.
Quote from: dioxin on March 20, 2012, 09:33:56 PM
QuoteBut as dave show, -3^2.5=-15.5884572681 is correct.
No it isn't.
-3^2.5 = -3^2 * -3^0.5
= 9 * ±1.7320508j
= ±15.58845j
There are 2 solutions, both are imaginary.
Paul.
dioxin,
Where did you learn that ?
As far as i know
-3^2.5 means this: -(3^2.5) so make what you want (3^2 * 3^0.5)
If you want to raise (-3) to (2.5) write (-3)^2.5
What do you mean if you write this: +9-3^2 ?
Rui,
a unary minus sign is the highest priority mathematical operator. It's not really an operator at all but an intrinsic part of the number, to be applied before any other operator is considered.
-3^2 means (-3)^2 not -(3^2)
-3 is not 3 subtracted from zero, it is a single entity "minus3", just as 12 is not two separate digits but a single number "twelve". You can't remove the - from the 3 without changing its meaning any more than you can remove the 2 from the 1 and still leave it as 12.
Neither can you convert the unary - to a binary operation by prefixing a zero which should not be there and then redo the calculation (which appears to be what you're doing):
0-3^2 = -9
as that 0 changes the meaning of the calculation.
Consider 2 * -3.
Do you not expect the result to be -6?
Or do you place a zero in the unoccupied position ahead of the minus sign to convert it to a binary operation and end up with
2*0-3 = -3
Excel gives -3^2 = 9
My 2 pocket calculators give -3^2 = 9
From what I was taught at school -3^2 = 9
Microsoft PowerToy calculator gives -3^2 = 9
Perhaps some programming languages have it wrong. If so then it's a good job they have () to circumvent the error.
Paul.
My bad. I realized my error while having dinner.
-3^2.5 = -15.5884572681 is entirely correct because the "^" takes precedence over the "-" sign in such an expression. The "-" must thus be used only after the power is applied. It is thus equal to -(3^2.5).
(-3)^2.5 correctly results in ERROR.
My apology again for my ?????. Your prog was correct.
Paul,
Your calculator gives you -3^2=9 because you effectively put backets around the -3 before applying the power, (-3)2'
Think of the expression 4*5-3^2.
You would first apply the "*" operator, then the "^" operator, and apply the "-" operator as the last one, certainly not before the "^" operator unless you use brackets such as (4*5-3)^2.
Raymond,
no implied brackets anywhere. The calculator takes the full algebraic expression and then calculates the result when I press the = key so it is judging precedence of the operators with unary - ahead of ^.
Perhaps it's an age thing. My main calculator is over 30 years old. Maths was done properly in those days!
Your second case is different because the - is a binary operator there.
In a case such as -3^2 it's clearly unary and is part of the number minus3.
With 4*5-3^2 it's clearly binary otherwise there is no operator between the 4*5 and the -3^2
You could have 4*5- -3^2 where the - acts as binary the first occurence and unary the second occurence.
It's not so obvious with something like:
x=-3
-x^2 = ?
In this case the - is not binary but neither is it clearly part of the number x. It's actually a negation operator on what follows but it makes sense to me, and it's how I was taught, that the unary minus is the highest precedence operator so, even in this case the - should be applied first giving -x^2 = 9 when x=-3.
There are differences of opinion out there. Programming languages C and BC, Sharp Corporation of Japan (the makers of my calculator), Microsoft, and My various maths teachers all put unary - right at the top in precedence. It seems more logical to do it that way otherwise the whole range of negative numbers are not representable except as a negation of a positive number and there's no reason to arbitrarily relegate half of all real numbers to be functions of the other half in that way. Maths is comlex enough without that sort of thing.
Paul.
Quote from: dioxin on March 21, 2012, 12:33:49 AMit's how I was taught, that the unary minus is the highest precedence operator so, even in this case the - should be applied first giving -x^2 = 9 when x=-3.
that assumption is false!
-x^2 = -1*x^2 = -1*(-3)^2 = -9
try this one: http://www.wolframalpha.com
Rui's program is correct
i was thinking -3^2.5 was complex :P
his program does return an error for (-3)^2.5
and -3^2.5 should be interpreted the same as -(3^2.5)
http://en.wikipedia.org/wiki/Order_of_operations
you'd think we all would know that, by now :bg
QuoteExcel gives -3^2 = 9
I use QuattroPro, Corel's version of Excel. When I enter -3^2, it displays -9 as the answer. Only when I enter (-3)^2 do I get 9 as the answer.
Entering -3^2.5, I get -15.58846 but entering (-3)^2.5, it results in ERROR.
I wonder how come those two similar programs would produce different results with the same input.
ok - gotcha
Excel seems to be messed up
OpenOffice also returns 9 for -3^2.
It also returns an error for -3^2.5!!!! :dazzled:
qWord,
not false at all. Given there are 2 ways to interpret it, I'd suggest that my way makes more sense.
To do it your way, the - in -2 becomes a negation operator on the number 2. It's then only possible to represent that negative number in terms of a function on the positive number. That makes no sense as there is no fundamental reason for negative numbers to be derived in this way. It's far more sensible to have -2 directly represent the required number in which case the - is not an operator but a part of the number.
If the - in -2 is to be treated this way then it follows that the - in -x ought to be treated the same otherwise we don't now have 2 uses of -, we have 3, one is a representation of a negative number, one as a binary subtraction operator and a one as a unary negation operator.
There is no reason to assume that -x is really -1 * x, that's fiddling the facts to match the desired outcome which just adds unnecessary complexity. You start adding a 1 after a - if it precedes a variable but not if it precedes a literal? You then must assume implied multiplication when programming languages never use implied multiplication. You then add 0 ahead of a - if there is no other number ahead of it (as in -x^2 should be treated as 0-x^2 where the unary - is changed to a binary -)?
Why do these fiddles when treating - as the highest priority removes the need for any fiddling as well as reducing complexity and making complete sense?
dedndave,
a quote (carefully framed to make a point!) from the Wikipedia page you refer to:
Quoteunary operators have a higher priority than binary operators, that is, the unary minus (negation) has higher precedence than exponentiation
Raymond,
OpenOffice Calc follows the Excel rules and gives -3^2 = 9
dedndave,
QuoteExcel seems to be messed up
no, you have it wrong. Excel got it right. Open Office agrees as do C, BC, Sharp, Mr.Duggan (my maths teacher) and others.
Paul.
Quote from: dioxin on March 21, 2012, 01:55:20 AM
Why do these fiddles when treating - as the highest priority removes the need for any fiddling as well as reducing complexity and making complete sense?
-3^2 = (-3)*(-3) = 3^2
-3^2 = 3^2 ???
make this sense?
Which all leads to the conclusion that, if you write an expression starting with a "-" sign, use brackets to clarify the expression. Otherwise, it risks being misinterpreted.
qWord,
Quotemake this sense?
Yes, perfect sense.
All real, positive numbers have 2 real square roots which differ only in sign.
sqrt(9) = +3 and -3 since (+3)*(+3) = 9 and (-3)*(-3) = 9.
Paul.
well - a little reading...
it appears there are "differing conventions" between mathematicians and programmers using various languages
so - Excel, C, etc - noone is "wrong", per se :P
the same can be said of the "/" symbol, for example, 1/2X may be read 2 ways
as Raymond pointed out, use parens if needed
i never realized there were "differing conventions"
still learn something new everyday - lol
good to know and be aware of
Quote from: dioxin on March 21, 2012, 02:30:45 AM
qWord,
Quotemake this sense?
Yes, perfect sense.
All real, positive numbers have 2 real square roots which differ only in sign.
indeed, this was a weak example.
However, for 'common' algebra, -x^2 is treated as -(x^2).
If there are calculator or solvers, that do not follow this, then they do this probably because of technically reasons (like my SmplMath-macros)
qWord
qWord,
Quotethen they do this probably because of technically reasons
The reason the calculator does it is because it was considered the correct way for it to be done, not because of some technicality which couldn't be avoided.
Paul.
Paul,
I have no time to discuss rules and logic problems.
I have a lot of things to do in the next few days.
In any way raymond told one general rule when we write
one expression:
«if you write an expression starting with a "-" sign,
use brackets to clarify the expression»
But you dont answer my question:
What do you mean if you write this: +9-3^2 ?
one note: 1. sqrt(9) = 3
2. If we are solving the equation x^2-9=0
the solutions are: x1=+3, and x2=-3
Quote
-3^2 means (-3)^2 not -(3^2)
Folowing your own rules, we have:
(a) -(+3)^2 = (-3)^2= (-3)*(-3)= 9
(b) -(-3)^2 = (+3)^2= (+3)*(+3)= 9
(c) +(+3)^2 = (+3)^2= (+3)*(+3)= 9
(d) +(-3)^2 = (-3)^2= (-3)*(-3)= 9
Its beautiful ! It seems we are in the Zeta space
one note more:
From Wikipedia you can read:
«Exceptions to the standard»
«There exist differing conventions concerning the unary operator -
(usually read "minus").
In written or printed mathematics, the expression -3^2 is interpreted to mean -(3^2) = -9,
...?»
I repeat to you (as i taught this things a lot of years ago)
In mathematics -3^2 is interpreted to mean -(3^2) = -9.
Rui,
QuoteBut you dont answer my question:
What do you mean if you write this: +9-3^2 ?
I did answer it. The - in that case is a binary operator and has the same precedence as +. It's the unary minus which causes problems.
QuoteFolowing your own rules, we have:
(a) -(+3)^2 = (-3)^2= (-3)*(-3)= 9
(b) -(-3)^2 = (+3)^2= (+3)*(+3)= 9
(c) +(+3)^2 = (+3)^2= (+3)*(+3)= 9
(d) +(-3)^2 = (-3)^2= (-3)*(-3)= 9
No we don't! As soon as you separate the - from the number that follows it then it ceases to be part of that number.
-3^2 = 9 'the - is part of the number minus3, it's not a separate entity from the digit 3
-(3)^2 = -9 'the - is not part of any number but becomes an operator on the result of the bracketed expression
So we actually get:
(a) -(+3)^2 = -(3)^2 = -(3*3) = -9
(b) -(-3)^2 = -(-3)^2 = -(-3*-3) = -9
(c) +(+3)^2 = (+3)^2 = (+3)*(+3) = 9
(d) +(-3)^2 = (-3)^ 2 = (-3)*(-3) = 9
Quoteone note more:
From Wikipedia you can read:
«Exceptions to the standard»
I think the one clear thing that has been established here is that there is no standard.
It's not just me versus everyone else. There are significant other sources which do it the way I say.
Paul
Paul,
1. Ok, i dont want to follow that rules of unary minus or whatever. :wink
2. I usually doesnt use Wikipedia. But as we can see there,
«In mathematics -3^2 is interpreted to mean -(3^2) = -9.»
3. What is the problem ? On your first post, you said:
«no it isn't». It means that your interpretation is
correct and the others are incorrect or dont exist.
It is a question of rules.
Quote
There are significant other sources which do it the way I say.
4. and significant other sources which doesnt follow it.
5. Well, what are the rules ?
This:
The order of operations, or precedence (
used throughout mathematics,
science, technology and many computer programming languages) is:
1. terms inside parentheses or brackets
2. exponents and roots
3. multiplication and division
4. addition and subtraction
I dont change that rules now. In this way
if we write 1+2^3*2/3, calcula20 start with 2^3 then *, /, +
More: it we type 2^3^4, calcula20 starts by the end: 3^4
qWord,
-3^2 = (-3)*(-3) = 3^2
-3^2 = 3^2 ???
make this sense?
No. To me it doesnt.
The same to -3^2+9 = 9+9 = 18
but 9-3^2 = 9-9 = 0 (we lost the commutativity)
If x=-3^2 stands for a real number and y=9 is another real number
then x+y=y+x (commutativity law).
RuiLoureiro,
Quote from: RuiLoureiro on March 21, 2012, 07:19:35 PM
The order of operations, or precedence (used throughout mathematics,
science, technology and many computer programming languages) is:
I'm right there with you.
qWord,
Ok ! :U :wink
Hi all !
Calcula21, now solves some more cases that calcula20 doesnt solve.
Remember that:
-X^Y means this -(X^Y)
Now, we have: e^x = EXP(x) = exp(x)
To get e=2.718... e^1
Good luck !
Did you note some errors ?
Please tell me !
Thanks
note: cases like: (-2)^(-5/3) ... (-2)^(2/3) ...
note: it can solve up to 200 operatios in each (...)
and we can write 200 (...)
Calculation rules
1. terms inside parentheses or brackets
2. functions
3. exponents or powers
4. multiplication and division
5. addition and subtraction
a) All calculatios are made in 80 bit format
e.g.: 2+3 both are taken as real10 numbers
b) Exceptions: expressions X^Y. Y is taken as integer32
if possible;
e.g.: (-2)^(2+3*1/3-1) but not here (-2)^(2+3*1/3-1.0)
because one term is 1.0
35 Kb EXE file
16 Kb of it is 0's :bg
you could use uninitialized data for a lot of it
but, alas, we cannot see the source :(
Quote from: dedndave on March 22, 2012, 07:27:47 PM
35 Kb EXE file
16 Kb of it is 0's :bg
you could use uninitialized data for a lot of it
but, alas, we cannot see the source :(
Dave,
read again the post.
May be i solve that problem in Calcula22 !
well - i couldn't find any problems with Calcula20 :P
Dave,
Quote
well - i couldn't find any problems with Calcula20
With calcula20:
---------------------------------------------
Type an expression and press enter [ ex: -pi + 3.1*(7 - 2*8.3) ]
(-2)^(3/5)
This is your expression: (-2)^(3/5)
This is your Result: ERROR
(-2)^(3/5)=ERROR
---------------------------------------------
Quote
35 Kb EXE file
16 Kb of it is 0's
Now calcula22 has 19.5K. I used uninitialized data.
I dont know if i alloc mem it isnt shorter
***
I found one
error in Calcula21.
It is solved in calcula22.
Thanks
EDIT:
Sorry, i failled using fcomip st(0),st(1)
and the result was allways 0 ! It is solved
The problem is to remove -0.00000