如何查看gdb内部智能指针的内部数据? [英] How to view the internal data of a smart pointer inside gdb?

问题描述

我有如下测试程序:

#include<memory>
#include<iostream>
using namespace std;

int main()
{
    shared_ptr<int> si(new int(5));
    return 0;
}

调试它:

(gdb) l
1   #include<memory>
2   #include<iostream>
3   using namespace std;
4   
5   int main()
6   {
7       shared_ptr<int> si(new int(5));
8       return 0;
9   }
10  
(gdb) b 8
Breakpoint 1 at 0x400bba: file testshare.cpp, line 8.
(gdb) r
Starting program: /home/x/cpp/x01/a.out 

Breakpoint 1, main () at testshare.cpp:8
8       return 0;
(gdb) p si
$1 = std::shared_ptr (count 1, weak 0) 0x614c20

它只打印出si的指针类型信息，但是如何获取其中存储的值(本例为5)?调试时如何查看si的内部内容?

It only prints out the pointer type information of si, but how to get the value stored in it (in this case 5)? How can I check the internal content of si during debugging?

推荐答案

试试以下:

p *si._M_ptr

现在，假设您正在使用 libstdc++.so，给出 p si 的输出.

Now, this assumes that you're using libstdc++.so, given the output for p si.

或者，您可以直接使用值 0x614c20(来自您的输出):

Alternatively, you could use the value 0x614c20 directly (from your output):

p {int}0x614c20

两者都应显示值 5.

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