Re: Single stepping a simple C-program, but...

[Pedro Alves <pedro@codesourcery.com>]

A Monday 19 May 2008 21:18:36, Peter Toft wrote:
> Hi guys
>
> I was a bit surprised today with GDB, and I hope one of you can explain it
> to me. Take a look at http://pastebin.org/37117

01. #include <stdio.h>       
02. 
03. int main(void)
04. {
05.   int ii=4;      
06.                         
07.   if ((ii>3) || (ii<1))  
08.    printf("hej A\n");     
09.   else                   
10.    printf("hej B\n");    
11.                         
12.   return 0;              
13. }

> Press download and save as my_program.c
> $ gcc -g my_program.c
> $ gdb ./a.out
> (gdb) br 7
> Breakpoint 1 at 0x804838c: file my_program.c, line 7.
> (gdb) r
> Starting program: /home/pto/c/a.out
>
> Breakpoint 1, main () at my_program.c:7
> 7	  if ((ii>3) || (ii<1))
> (gdb) s
> 8	    printf("hej A\n");
> (gdb) s
> hej A
> 7	  if ((ii>3) || (ii<1))     <----------- WHY!!!!??
> (gdb) s
> 12	  return 0;
>
> -------
>
> Why does the second "step" i.e. "s" take me BACK to line 7 after I
> have been in line 8????
>

Presumably, because there's a branch instruction after the printf
call to skip the else clause, GDB hits it (step-resume breakpoint),
and the debug info gcc is producing is marking that address as belonging
to line 7?

-- 
Pedro Alves