this is a simple Assembly program. I am trying to output the Emp1 And Num1 in the following format:
Jow Blow
600 763 521
i get the following output:
Jow Blow
x
Also i want to output the total of the three numbers
.386
.MODEL FLAT
ExitProcess PROTO NEAR32 stdcall, dwExitCode:DWORD
Include io.h
.STACK 4096 ; reserve 4096-byte stack
.DATA ; reserve storage for data
Emp1 byte 'Jow Blow',13,10,0
Num1 DWORD 600,763,521
.CODE ; start of main program code
_start:
mov eax, Num1 ; first number to EAX
add eax, 158 ; add 158
mov sum, eax ; sum to memory
output Emp1
output Num1
INVOKE ExitProcess, 0 ; exit with return code 0
PUBLIC _start ; make entry point public
END ; end of source code
Quote600 763 521
Also i want to output the total of the three numbers
;
;
Num1 DWORD 600,763,521
probably the simplest way to handle this is to split Num1 into 3 parts
you could also do it by using a bunch of code to split it up (dividing by 1000 twice)
the method you choose would depend on how many such numbers you will have
for one, or a few entries, i would split it up...
Num1a dd 600
Num1b dd 763
Num1c dd 521"dd" is a shortcut - same as DWORD
Thank you for your help dedndave
but is there a way to do it without splitting it
Num1 DWORD 600,763,521
is the same as
Num1a dd 600
Num1b dd 763
Num1c dd 521
The difference is, that the second form allows access through names, while the first version would require something like this:
mov eax,Num1[0]
mov edx,Num1[4]
mov ecx,Num1[8]
You must convert the result sum into a string. The common method is to divide the number trough the radix (e.g. 10) until the result gets zero. The division's reminder are summed up with 0x30 (='0') and concatenated to a string, which can be printed.
Thank you qWord
.DATA
Num1 dd 600763521
.DATA?
PartA dd ?
PartB dd ?
PartC dd ?
.CODE
start: mov ecx,1000
mov eax,Num1
mov edx,0
div ecx
;EAX = 600763
;EDX = 521
mov PartC,edx
mov edx,0
div ecx
;EAX = 600
;EDX = 763
mov PartB,edx
mov PartA,eax