Signal Handler question

ntubski · 04-13-2022, 07:51 PM

Instead of guessing, you can check with objdump -S

Code:

int main()
{
   0:   55                      push   %rbp
   1:   48 89 e5                mov    %rsp,%rbp
        signal(SIGFPE, &funct);
   4:   48 8d 35 00 00 00 00    lea    0x0(%rip),%rsi        # b <main+0xb>
   b:   bf 08 00 00 00          mov    $0x8,%edi
  10:   e8 00 00 00 00          callq  15 <main+0x15>
        a = 1/0;//loads to register R2 therefore I have 1/0 in R2
  15:   b8 01 00 00 00          mov    $0x1,%eax
  1a:   b9 00 00 00 00          mov    $0x0,%ecx
  1f:   99                      cltd
  20:   f7 f9                   idiv   %ecx
  22:   89 05 00 00 00 00       mov    %eax,0x0(%rip)        # 28 <main+0x28>
        printf("ddd\n");
  28:   48 8d 3d 00 00 00 00    lea    0x0(%rip),%rdi        # 2f <main+0x2f>
  2f:   e8 00 00 00 00          callq  34 <main+0x34>
        return 0;
  34:   b8 00 00 00 00          mov    $0x0,%eax
}
  39:   5d                      pop    %rbp
  3a:   c3                      retq

000000000000003b <funct>:

void funct()
{
  3b:   55                      push   %rbp
  3c:   48 89 e5                mov    %rsp,%rbp
        printf("done\n");
  3f:   48 8d 3d 00 00 00 00    lea    0x0(%rip),%rdi        # 46 <funct+0xb>
  46:   e8 00 00 00 00          callq  4b <funct+0x10>
        a=900;//does not matter if I have a=900,I still have 1/0 in R2
  4b:   c7 05 00 00 00 00 84    movl   $0x384,0x0(%rip)        # 55 <funct+0x1a>
  52:   03 00 00
        exit(0);
  55:   bf 00 00 00 00          mov    $0x0,%edi
  5a:   e8 00 00 00 00          callq  5f <funct+0x24>

pan64 · 04-13-2022, 11:41 PM

Quote:

Originally Posted by ntubski

Instead of guessing, you can check with objdump -S

That is fine, just we can't see how does it work.
the line

Code:

20:   f7 f9                   idiv   %ecx

will cause that sigfpe, and we would need to use a debugger, set a breakpoint in the signal handler and check where will it return to (if the exit statement is missing). Or we can print the stack (without debugger). Anyway, it will return to the same line and will [try to] execute it again.

I did not find relevant documents about it, so probably I'm wrong.

sundialsvcs · 04-15-2022, 01:46 PM

You really don't have to "dumpster-dive into assembly code" to understand the essential idea. If you are handed a SIGSEGV, your process probably cannot reasonably continue. So, the signal handler orders the process to exit.