Why watchpoint doesn't effect?

Question

I am studying the watchpoint of GDB. I write a simple test code as following:

int main(int argc, char **argv)
{
    int x = 30;
    int y = 10;

    x = y;

    return 0;
}

I build it via gcc -g -o wt watch.c. And then I started gdb and did following experiment:
    lihacker@lihacker-laptop:~/mySrc$ gdb ./wt
    GNU gdb (GDB) 7.3
    Copyright (C) 2011 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-linux-gnu".
    For bug reporting instructions, please see:
    <http://www.gnu.org/software/gdb/bugs/>...
    Reading symbols from /home/lihacker/mySrc/wt...done.
    (gdb) b main
    Breakpoint 1 at 0x80483a5: file watch.c, line 5.
    (gdb) run
    Starting program: /home/lihacker/mySrc/wt 

    Breakpoint 1, main (argc=<optimized out>, argv=<optimized out>) at watch.c:5
    5     int x = 30;
    (gdb) watch x
    Hardware watchpoint 2: x
    (gdb) c
    Continuing.

    Watchpoint 2 deleted because the program has left the block in
    which its expression is valid.
    0xb7e83775 in __libc_start_main () from /lib/tls/i686/cmov/libc.so.6
    (gdb)

In my test codes, the variable "x" is changed, but gdb doesn't stop then. Why the watchpoint doesn't effect here? Thanks a lot.

score 1 · Answer 1 · answered Dec 26 '11 at 16:27

This:

Breakpoint 1, main (argc=<optimized out>, argv=<optimized out>) at watch.c:5

suggests that you used -O2 or some such flag when building the test. Try building with -O0 (which will explicitly disable optimization).

Even then, there is a glitch (a buglet) in GDB. Here is what I see:

(gdb) b main
Breakpoint 3 at 0x80483ba: file t.c, line 3.
(gdb) r

Breakpoint 3, main (argc=1, argv=0xffffca94) at t.c:3
3       int x = 30;
(gdb) watch x
Hardware watchpoint 4: x
(gdb) c
Hardware watchpoint 4: x

Old value = 0
New value = 10
main (argc=1, argv=0xffffca94) at t.c:8
8       return 0;
(gdb) c

Watchpoint 4 deleted because the program has left the block in
which its expression is valid.
0xf7e7cbd6 in __libc_start_main () from /lib32/libc.so.6

This can't be right: the value of x changes from 30 to 10, not from 0 to 10.

If I set the breakpoint on the very first instruction of main, then it works as expected:

(gdb) b *main
Breakpoint 1 at 0x80483b4: file t.c, line 2.
(gdb) r

Breakpoint 1, main (argc=1, argv=0xffffca94) at t.c:2
2   {
(gdb) watch x
Hardware watchpoint 2: x
(gdb) c
Hardware watchpoint 2: x

Old value = 0
New value = 30
main (argc=1, argv=0xffffca94) at t.c:4
4       int y = 10;
(gdb) c
Hardware watchpoint 2: x

Old value = 30
New value = 10
main (argc=1, argv=0xffffca94) at t.c:8
8       return 0;
(gdb) c

Watchpoint 2 deleted because the program has left the block in
which its expression is valid.
0xf7e7cbd6 in __libc_start_main () from /lib32/libc.so.6

Thanks for your reply. Actually I didn't set any flag for optimization when building via gcc. The result is the same even though I add -O0 as you said. My test platform is Ubuntu 9.0.4 which is running on a Virtual machine and my gdb version is latest 7.3. What's your test environment? — mingganz, Dec 28 '11 at 04:45

Why watchpoint doesn't effect?

1 Answers1