gdb中C ++引用修改的硬件断点

时间:2017-11-10 10:13:00

标签: c++ gcc gdb

C ++标准说it is unspecified whether or not a reference requires storage (3.7).。但是,据我所知,gcc将C ++引用实现为指针,因此它们可能会被破坏。

是否有可能在gdb中获取引用的地址并在该地址上放置硬件断点,以便找出损坏引用所在内存的内容?如何设置这样一个断点?

1 个答案:

答案 0 :(得分:0)

GDB可能会进行硬件监视。您可以使用命令watch。例: 的 main.cpp中:

int main(int argc, char **argv)
{
     int a = 0;
     int& b = a;
     int* c = &a;
     *c = 1;

     return 0;
}

启动调试并在启动主函数和结束主函数上设置断点:

(gdb) b main
Breakpoint 1 at 0x401bc8: file /../main.cpp, line 60.
(gdb) b main.cpp:65
Breakpoint 2 at 0x401be9: file /../main.cpp, line 65.
(gdb) r

获取参考b的地址:

Breakpoint 1, main (argc=1, argv=0x7fffffffddd8) at /../main.cpp:60
60           int a = 0;
(gdb) disas /m
Dump of assembler code for function main(int, char**):
59      {
   ... Something code

60           int a = 0;
=> 0x0000000000401bc8 <+11>:    movl   $0x0,-0x14(%rbp)

61           int& b = a;
   0x0000000000401bcf <+18>:    lea    -0x14(%rbp),%rax
   0x0000000000401bd3 <+22>:    mov    %rax,-0x10(%rbp)

62           int* c = &a;
   0x0000000000401bd7 <+26>:    lea    -0x14(%rbp),%rax
   0x0000000000401bdb <+30>:    mov    %rax,-0x8(%rbp)

63           *c = 1;
   0x0000000000401bdf <+34>:    mov    -0x8(%rbp),%rax
   0x0000000000401be3 <+38>:    movl   $0x1,(%rax)

64      
65           return 0;
   0x0000000000401be9 <+44>:    mov    $0x0,%eax

66      }
   0x0000000000401bee <+49>:    pop    %rbp
   0x0000000000401bef <+50>:    retq   

End of assembler dump.
(gdb) p $rbp-0x10
$1 = (void *) 0x7fffffffdce0

p $rbp-0x10是参考b的打印地址。它是0x7fffffffdce0。 设置此地址以供观看:

(gdb) watch *0x7fffffffdce0
Hardware watchpoint 3: *0x7fffffffdce0
(gdb) c

只有在值发生变化时,GDB才会中断:

(gdb) c
Continuing.
Hardware watchpoint 3: *0x7fffffffdce0

Old value = -8752
New value = -8996
main (argc=1, argv=0x7fffffffddd8) at /../main.cpp:62
62           int* c = &a;

抱歉我的英文!