gdb“watch”不能通过glibc（read）函数修改变量？

Question

当chmodified.I usedwatch chin gdb，我想打破它，它不起作用。

像ch=1;这样的东西会坏掉。为什么read()不呢？

这样使用watch 命令是正确的。或者read()函数是特殊的？

对不起我的英语，代码说明一切。

文件 1.c：

#include <unistd.h> 
#include <sys/types.h> 
#include <sys/stat.h> 
#include <fcntl.h> 
#include <stdio.h> 
const char *const filename = "/etc/passwd"; 
int main(void) 
{ 
     int fd; 
     int ch; 
     fd = open(filename, O_RDONLY); 
     read(fd, &ch, sizeof(int));  
     printf ("%d\n", ch); 
     close (fd); 
     return 0; 
}

gcc -g 1.c

调试：

$ gdb a.out  
 GNU gdb (GDB) 7.4.1-debian 
 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 "i486-linux-gnu". 
 For bug reporting instructions, please see: 
 <http://www.gnu.org/software/gdb/bugs/>... 
 Reading symbols from /home/zodiac1111/tmp/a.out...done. 
 (gdb) b main 
 Breakpoint 1 at 0x80484b5: file 1.c, line 11. 
 (gdb) r 
 Starting program: /home/zodiac1111/tmp/a.out  

 Breakpoint 1, main () at 1.c:11 
 11        fd = open(filename, O_RDONLY); 
 (gdb) watch ch 
 Hardware watchpoint 2: ch 
 (gdb) c 
 Continuing. 
 1953460082 

 Watchpoint 2 deleted because the program has left the block in
 which its expression is valid. 
 __libc_start_main (main=0x80484ac <main>, argc=1, ubp_av=0xbffff4c4,  
     init=0x8048530 <__libc_csu_init>, fini=0x8048520 <__libc_csu_fini>,  
     rtld_fini=0xb7ff0590, stack_end=0xbffff4bc) at libc-start.c:260 
 260    libc-start.c: No such dir...
 (gdb) c 
 Continuing. 
 [Inferior 1 (process 9513) exited normally]

Answer 1

对于read() 的正常实现，对内存的写入将由内核直接执行，而不是由任何用户空间代码执行。调试器没有在内核中放置断点的机制，即使有，它也无权这样做。