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我想在ch修改时中断。我watch ch在 gdb 中使用过，它不起作用。

像ch=1;会坏的东西。为什么read()不呢？

是正确的使用watch这样的命令。还是read()功能特殊？

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]
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1 回答 1

4

对于 的正常实现read()，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，它也没有这样做的权限。

于 2013-09-24T06:10:29.140 回答