hi
how would i use Print to output a number in binary.
thanks
Rainstorm
Print(1);
Print(0);
Print(0);
Print(1);
Print(0);
Print(1);
Print(0);
Print(1);
:bdg
Or you could convert the number to a string of binary digits first, null-terminate it, and then print that :wink
lol
this is the code i had tried, but that gives it in decimal for the byte in 'al'
how would i output/display the byte in al, in binary
start:
xor eax, eax
mov al, 11001b
And al, al
movzx eax, al
print ustr$(eax),13,10
inkey
exit
i tried with byt2bin_ex it assembles proper..but get the windows error msg.
.data
buff10 db 12 (0)
pbuff dd 0
var10 db 0
.code
start:
xor eax, eax ; clear eax
mov al, 10001b
And al, 1 ; some random operation
mov ebx, offset buff10
mov pbuff, ebx
;mov var10, al
invoke byt2bin_ex,al,pbuff
print pbuff,13,10
inkey
exit
end start
thnks..
loop..
- get each bit by shifting (or rotating)
- test if a bit is 1, then "setnz al"
- add al,'0'
- al is now the "0" or "1" to print
remember to get the bits 'backwards' (so you print the highest bit first)
ted,
i can't get how to do it like you said...
after shifting..the bit gets stored in the CF flag right ?
i attempted to do something like in the following code..
but it doesn't seem to be outputing the right results..& i don't know why.
also do you know why my previouss code with the byt2bin_ex function isn't working ?
.data
buff db 64 dup(0)
pbuff dd 0
.code
start:
xor bl, bl
xor al, al
mov bl, 00011001b
sal bl, 1
setnc al
setc al
mov buff,al
mov esi, offset buff
mov pbuff, esi
print pbuff,13,10
inkey
exit
end start
thanks..
hi,..
here i tried some code on the lines of your post..- it assembles proper.. but doesn't give the
right results,
Also, in your post i don't know what add al,'0' does exactly
.data
buff db 64 dup(0)
pbuff dd 0
msk db 0
.code
start:
xor bl, bl
xor al, al
mov esi, 0
mov bl, 0
mov al, 10000000b
printt:
mov msk, 00011001b
test msk, al
setnz cl
movzx esi, bl
mov buff[esi], cl
mov buff[esi+1], 0
mov edi, offset buff
push eax
print edi
pop eax
sar al, 1
inc esi
inc bl
cmp bl, 7
jz xit
jmp printt
xit:
print " ",10
inkey
exit
end start
Quote from: Rainstorm on November 22, 2006, 11:27:26 PM
do you know why my previouss code with the byt2bin_ex function isn't working ?
Because you passed the value 'buff' ( = 0) instead of the address. Try this:
invoke byt2bin_ex,al,OFFSET pbuffCheers,
Zooba :U
zooba,
QuoteBecause you passed the value 'buff' ( = 0) instead of the address. Try this:..
not understanding how i passed the value as 0..
mov ebx, offset buff10
mov pbuff, ebx ; that moves the address of the buffer into pbuff...right ?
invoke byt2bin_ex,al,pbuffthanks..
You're right, sorry. I didn't read the code properly :red
Rainstorm,
Give this a blast.
; «««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««
include \masm32\include\masm32rt.inc
; «««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««
comment * -----------------------------------------------------
Build this template with
"CONSOLE ASSEMBLE AND LINK"
----------------------------------------------------- *
.data?
byte_buffer db 32 dup (?)
.code
start:
; «««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««
call main
inkey
exit
; «««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««
main proc
LOCAL pbuf :DWORD
LOCAL buffer[32]:BYTE
push ebx
; ----------------
; use LOCAL memory
; ----------------
xor ebx, ebx ; set EBX to zero
mov pbuf, ptr$(buffer)
@@:
invoke byt2bin_ex,bl,pbuf ; use BL as the BYTE value
print pbuf,13,10 ; display the result
add ebx, 1 ; increment EBX for next BYTE value
cmp ebx, 256 ; exit if > 255
jne @B
inkey ; wait for a keypress
; ---------------------------------------
; use GLOBAL memory in the .DATA? section
; ---------------------------------------
xor ebx, ebx ; set EBX to zero
mov pbuf, OFFSET byte_buffer ; load the OFFSET into a pointer
@@:
invoke byt2bin_ex,bl,pbuf ; use BL as the BYTE value
print pbuf,13,10 ; display the result
add ebx, 1 ; increment EBX for next BYTE value
cmp ebx, 256 ; exit if > 255
jne @B
pop ebx
ret
main endp
; «««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««««
end start
hutch,..
thanks a lot.
Rainstorm
don't know the reason, my old code using byt2bin_ex didn't work though.
was trying to compare it with yours..
-
Another way.
.data
Src dd 87654321h, 0 ; could be used instead of array of bytes to show dwords
szSrc db 21h,43h,65h,87h,0,0,0,0 ; in standard lo byte first format (little endian)
szCap db "Binary output:",13,10
szMsg db 256 dup (0)
.code
start:
xor edi, edi
mov esi, 4 ; count of bytes
NB:
mov ecx, 8
movzx eax, byte ptr [szSrc+esi-1]
@@:
shl eax, 1
xor eax, 3000h
mov byte ptr [szMsg+edi], ah
inc edi
and eax, 0FFh
dec ecx
jnz @B
mov byte ptr [szMsg+edi], ' ' ; remove, only in to highlight byte boundaries
inc edi ; remove, only in to highlight byte boundaries
dec esi
jnz NB
mov byte ptr [szMsg+edi], 'b' ; remove, only have trailing b to indicate binary number
print offset szCap,13,10
inkey
exit
end start
Edit added the missing offset needed for the print macro, also added a : in caption string.
Using Hutch's nice example, I notice that using global memory is faster than using local memory. An interesting observation.
Paul
Swap the order around and now local memory is faster than global memory. It's probably just quicker the second time around :U
dsouza123 wrote
QuoteAnother way...
was curious to see how it was done that way too..
just digesting all that you posted..
thanks..
:U