title: Why watchpoint doesn't effect?
body: <p>I am studying the watchpoint of GDB. I write a simple test code as following:</p>
  
  <pre><code>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 &lt;http://gnu.org/licenses/gpl.html&gt;
      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:
      &lt;http://www.gnu.org/software/gdb/bugs/&gt;...
      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=&lt;optimized out&gt;, argv=&lt;optimized out&gt;) 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) 
  </code></pre>
  
  <p>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.</p>