CODEWITHSUNDEEP
c++linuxprintinggdbshared-ptr
Troskyvs
Troskyvs

Reputation: 8077

How to view the internal data of a smart pointer inside gdb?

I've got test program like below:

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

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

Debug it:

(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

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?

Upvotes: 9

Views: 9742

Answers (2)

ks1322
ks1322

Reputation: 35815

but how to get the value stored in it

You will have to cast raw pointer to actual pointer type stored in std::shared_ptr. Use whatis to know what the actual pointer type is.

(gdb) p si
$8 = std::shared_ptr (count 1, weak 0) 0x614c20
(gdb) whatis si
type = std::shared_ptr<int>
(gdb) p *(int*)0x614c20
$9 = 5

Upvotes: 3

geipel
geipel

Reputation: 405

Try the following:

p *si._M_ptr

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

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

p {int}0x614c20

Both should display the value 5.

Upvotes: 7

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