9
votes

System

Fresh install of codeblocks 12.11 + mingw pack.

  • win7 64
  • gcc 4.7.1
  • gdb 7.5

Example code

Compiled with -g and no optimization.

#include <stdio.h>

void foo(int a, int b);

int main() {

    foo(400, 42);

    return 0;
}

void foo(int a, int b) {

    a = a - 10;
    b = a + 1;

    printf("y2 %d\n", b);

}

Problem

I put a breakpoint on "void foo(int a, int b)" and I look value of b as I step through the 3 lines. Either with the codeblocks debugging features or with the gdb command line, the value of b is 42 instead of being 391. The console output is correct, 391.

GDB commands

C:\DebugTest>"C:\Program Files (x86)\CodeBlocks\MinGW\bin\gdb.exe"
GNU gdb (GDB) 7.5
Copyright (C) 2012 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.  Type "show copying"
and "show warranty" for details.
This GDB was configured as "i686-pc-mingw32".
For bug reporting instructions, please see:  
<http://www.gnu.org/software/gdb/bugs/>.
(gdb) file bin/Debug/DebugTest.exe
Reading symbols from C:\DebugTest\bin\Debug\DebugTest.exe...done.
(gdb) break foo
Breakpoint 1 at 0x401363: file C:\DebugTest\main.c, line 14.
(gdb) run
Starting program: C:\DebugTest\bin\Debug\DebugTest.exe
[New Thread 3596.0x658]

Breakpoint 1, foo (a=400, b=42) at C:\DebugTest\main.c:14
14          a = a - 10;
(gdb) print b
$1 = 42
(gdb) step
15          b = a + 1;
(gdb) print b
$2 = 42
(gdb) step
17          printf("y2 %d\n", b);
(gdb) print b
$3 = 42
(gdb)

Remarks

  • When the same operations are done without a function, with a and b as local variables inside main, the debug output is correct.
  • When compiled with gcc 4.4.1, the debug output is correct.

Any idea what could be wrong ? =)

3
I'm using Code Blocks 10.05 (GNU gdb 6.8), shows b as 391.SS Hegde
Are you printing the value of b in gdb after assignment b = a + 1;?ks1322
what do you mean with step inside foo? break foo, run then print b?Davide Berra
Yeah I print b both before and after b = a + 1 line. @DavideBerra I put a breakpoint on the void foo(int a, int b) { using codeblocks, I don't use gdb command except print. Updated the post.Teybeo
A complete set of gdb commands showing the problem could confirm or disprove some of those suspicions that it could just be a misuse/misunderstanding...aschepler

3 Answers

8
votes

I searched on gcc bugzilla and found this bug report :

Althoug the report is about gcc 4.8 and I'm using 4.7, I tried the proposed workaround and it works !

Compiling with -fvar-tracking allows GDB to print the correct value for b after assignment.

1
votes

gcc does not generate debugging info correctly for values that are in registers -- either values that have been put in registers or values that start there due to the calling conventions. This is a long-standing problem with gcc since at least 4.0, and makes it tricky to debug things.

0
votes

Sometimes the optimizer is smarter than the debugger. Try debugging unoptimized code, or step through disassembly and watch the HW registers directly rather than stepping through C source lines and watching the debugger's idea of what the variables are.