why is segmention fault while printing?

问题

This is my x86 assembly code:

section .data 
  output db '%d',10,0
section .text 
  global main
  extern printf 
main :
  xor ecx,ecx
  xor eax,eax
  mov eax,1
  mov ecx,5
lable1:
  push ecx 
  push eax  
  cmp eax,0
  jg print
  pop eax
  pop ecx
  inc eax
loop lable1 
  ret 
print:
  push eax
  push output
  call printf 
  add esp,8
  ret 

This program should print all numbers between 1 to 5. Why am I getting a segmentation fault after printing '1'?

回答1:

print ends with a ret instruction, which implies that it is something that you should call. So jg print should be jng skip / call print / skip: (or just call print, because the > 0 check seems unnecessary).
call places the return address on the stack, jg does not.

回答2:

The reason why you keep getting faults is: You are ignoring the stack!

Your lable1 subroutine is correctly set up, but with your jg you eject from it ignoring the stack - leaving the stack corrupt.

lable1:
  push ecx 
  push eax  
  cmp eax,0
  jg print
  pop eax
  pop ecx
  inc eax

You

xor eax,eax     ; EAX = 0 - breaking pipeline
mov eax,1       ; EAX = 1 - which is redundant

but because of your

cmp eax,0
jg print

in the lable1 subroutine, you jump to the print: routine if EAX is greater than 0(and it is in the first iteration with EAX=1) with a stack offset of -8 caused by

push ecx 
push eax  

in the beginning of lable1:. The retting from print:, which cleans up the stack properly by add esp,8 at the end, will cause your program to return to the address of the first entry of the stack, which you have assigned to [ESP]=(former EAX = 1) in main:mov eax,1.

Therefore you get a SegFault, because you try to jump/return to the address [00000001] (=EAX=1,32bit).

来源：https://stackoverflow.com/questions/36498852/why-is-segmention-fault-while-printing