更改环0中观察点的值会导致gdb的奇怪行为

Question

src.c：

main()
{
   char c='x';
   read(0,&c,1);
   printf("go\n");
   printf("come\n");
}

gcc src.c -g，你得到一个可执行文件a.out，然后， gdb a.out

(gdb) b main
Breakpoint 1 at 0x80483e5: file gdb.c, line 3.
(gdb) r
Starting program: /tmp/tx

Breakpoint 1, main () at gdb.c:3
3       char c='x';
(gdb) set $foo=&c
(gdb) watch *$foo   //set a watchpoint for the address in which c reside
Hardware watchpoint 2: *$foo
(gdb) del 1
(gdb) c
Continuing.
Hardware watchpoint 2: *$foo

Old value = 0 '\0'
New value = 120 'x'
main () at gdb.c:4
4       read(0,&c,1);
(gdb) c
Continuing.       **//type carrige return**

Hardware watchpoint 2: *$foo

Old value = 120 'x'
New value = 10 '\n'
0x00ae3402 in __kernel_vsyscall ()
(gdb)
Continuing.

Program received signal SIGTRAP, Trace/breakpoint trap. **//from now on,gdb receive SIGTRAP repeatedly,why?**

0x009f23f3 in __read_nocancel () from /lib/libc.so.6
(gdb)
Continuing.

Program received signal SIGTRAP, Trace/breakpoint trap.
0x009f23f4 in __read_nocancel () from /lib/libc.so.6
(gdb)
Continuing.

Program received signal SIGTRAP, Trace/breakpoint trap.
0x009f23f9 in __read_nocancel () from /lib/libc.so.6
(gdb)

但是，如果我将源代码修改为：

main()
{
    char c='x';

    printf("go\n");
    c++;

    printf("come\n");
}

然后一切顺利，没有有线行为。

在第一个程序中，watchpoint的值在内核模式下被修改，然后gdb接收SIGTRAP;在第二个程序中，watchpoint在用户模式下被修改，没有任何连接发生。有人告诉我为什么？提前谢谢。