为什么观察点不起作用？

Question

我正在研究GDB的观察点。我编写了一个简单的测试代码如下：

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) 

在我的测试代码中，变量“x”被更改，但gdb并没有停止。 为什么观察点在这里不起作用？多谢。

Answer 1

这：

断点 1，main（argc=<优化输出>、argv=<优化输出>）位于 watch.c:5

建议您使用 -O2 或类似的内容构建测试时进行标记。尝试使用 -O0 进行构建（这将显式禁用优化）。

即使如此，GDB 中仍然存在一个故障（buglet）。这是我所看到的：

(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

这不可能是正确的：x 的值从 30 更改为 10，而不是从 0 更改为 10。

如果我在 main 的第一条指令上设置断点，那么它按预期工作：

(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