I have a loop that runs succesfully 3 times, but the input I have in this loop works only the first time. I am new to assembly so pls have patience.
Code:
section .bss| d times 3 resb 1 ;array to store input| i resb 1 ;countersection .text| global _start|| _start:| | mov ecx, 3| | mov [i], byte 0| || | loop_here:| | | push ecx| | || | | mov eax, 3| | | mov ebx, 0| | | mov ecx, d| | | add ecx, [i]| | | inc byte [i]| | | mov edx, 1| | | int 80h| | || | | pop ecx| | loop loop_here| || | mov eax, 1| | xor ebx, ebx| | int 80hOutput:
2 ;I inserted 2 as an input2 ;I inserted 2 again as input[Program finishes]Well, later I thought that the loop might not running a third time, so I changed the code a bit.
New Code:
section .bss| d times 3 resb 1 ;array to store input| i resb 1 ;countersection .text| global _start|| _start:| | mov ecx, 3| | mov [i], byte 0| || | loop_here:| | | push ecx| | || | | mov eax, 4| | | mov ebx, 1| | | mov ecx, i| | | add [ecx], byte 30h ;to actually print the count| | | mov edx, 1| | | int 80h| | | sub [i], byte 30h ;to make again a counter| | | | | | mov eax, 3| | | mov ebx, 0| | | mov ecx, d| | | add ecx, [i]| | | inc byte [i]| | | mov edx, 1| | | int 80h| | || | | pop ecx| | loop loop_here| || | mov eax, 1| | xor ebx, ebx| | int 80hNew Output:
02122[Program finishes]Explaination:
0 is the counter and 2 my input. 1 and 2 are the counter and the second 2 is my second input, which means that the loop runs succesfully, it just ignores the code for input the second time it runs
Also, do the straight lines I have in my code to visualize the scopes make my code any more readable?
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Answer
Peter already did a great job of covering most of this in the comments, but I thought I’d go into more detail:
The x86 loop instruction assumes you have a counter in ecx. loop automatically decrements ecx and jumps to the loop label if ecx is not 0. Since you’re calling linux syscalls in a loop, and linux expects the buffer pointer to be in ecx, you should probably use a different register as your loop counter and some sort of jump instruction to do this instead:
mov esi, 3 ; esi is your loop counter - loop 3 timesloop_here:; do syscalldec esi ; decrement loop counterjnz loop_here ; jump to loop_here if esi is not zeroHowever, calling syscalls in a loop is not very efficient. Instead, you could do something like this:
mov eax, 3 ; readmov ebx, 0 ; fd for stdinmov ecx, d ; address of d buffer into ecxmov edx, 3 ; read 3 characters at most, the size of your bufferint 80hmov esi, eax ; read returns the number of bytes read in eax. ; we'll save it in esixor edi, edi ; edi will be our loop counter, this makes it 0loop_here: ; first, we'll print the loop counter valuemov eax, 4 ; writemov ebx, 1 ; fd for stdoutmov ecx, i ; address of i for writemov edx, 1 ; write 1 byteadd edi, '0' ; convert edi loop counter to ASCIImov [i], byte edi ; put lower byte of edi in iint 80hsub edi, '0' ; restore edimov eax, 4mov ebx, 1mov ecx, d ; address of d for write, add ecx, edi ; plus current loop counter (offset)mov edx, 1int 80hinc edi ; increment loop countercmp esi, edi ; compare to number of bytes readjne loop_here ; jump to loop_here if not equalThis calls read once, but still calls write in a loop to write the counter and values of d.
To skip the trailing newline in your input, you could try doing this after your call to read (remember eax has the number of bytes read):
movzx edi, byte [d+eax-1] ; move and zero-extend last byte of d into edicmp edi, `n` ; is it a newline? (backticks required)jne skip ; skip if notdec eax ; otherwise, we'll print one less byteskip:; rest of your code here The newline will still be stored in memory, but you won’t print it, because you’ll loop one fewer times.