How to access target of std::tr1::shared_ptr in GDB

Question

How can I access target of a std::tr1::shared_ptr in GDB. This doesn't work:

(gdb) p sharedPtr->variableOfTarget

If I try with the pointer object itself (p sharedPtr) I get something like this:

$1 = std::tr1::shared_ptr (count 2) 0x13c2060

With a normal pointer I can do p *ptr and get all the data or p ptr->variable for just one variable.

I'm on Centos 6.5, GCC 4.4.7-4.el6 and GDB 7.2-64.el6_5.2.

Which compiler and stdlib are you using? Also which version of GDB? — Alastair, Jul 23 '14 at 17:58
@Raydel Miranda That worked! Thanks very much! I'll add answer in a moment. — Pnog is not Pong, Jul 23 '14 at 18:30
I wasn't able to answer my own question yet, since I have too little reputation. This is what worked: `p (*sharedPtr.get())` prints contents of the target object and `p (*sharedPtr.get())->variableOfTarget` will print a single variable. Funny thing is though that if you leave out the parentheses like this `p *sharedPtr.get()` it will also advance execution of the program. Can someone explain why is that? — Pnog is not Pong, Jul 23 '14 at 19:07
`unique_ptr`: https://stackoverflow.com/questions/22798601/how-to-debug-c11-code-with-unique-ptr-in-ddd-or-gdb — Ciro Santilli OurBigBook.com, Sep 24 '17 at 12:00

score 34 · Answer 1 · answered Dec 22 '14 at 08:20

34

ptr->get() not always work.

when i try ptr->get(), gdb complains for: can not resolve method ***:get() to any overloaded instance

I eventually go to /usr/include/ to find the source code of shared_ptr to see the private member.

It turns out to be

ptr._M_ptr

It works for me. Source code works for everyone.

answered Dec 22 '14 at 08:20

boydc2011

567
1
5
7

You meant ptr.get() right? I think the operator `->` is not what you want use here. – Raydel Miranda Feb 09 '15 at 23:38
ptr->get() means *ptr.get(). – boydc2011 Feb 27 '15 at 03:27
3

Nop, is not the same, `ptr->get()` is an attemp to call the `get` function from whatever the shared_ptr is pointing to, and `*ptr.get()` is dereferencing that pointer. The `.` operator has preference here. `*ptr.get()` is the same that `*(ptr.get())`. – Raydel Miranda Feb 27 '15 at 12:59

Raydel Miranda · Accepted Answer · 2014-07-23T19:24:55.440

12

Try with

(gdb) p (*sharedPtr.get())

that function returns the a pointer to the object owned by the smart pointer.

edited Jul 23 '14 at 19:24

answered Jul 23 '14 at 18:53

Raydel Miranda

13,825
3
38
60

You have to use asterisk in front to get access to the pointer target and for some reason also parentheses like this: `(*sharedPtr.get())`. See my comment to the question. – Pnog is not Pong Jul 23 '14 at 19:12
If I use this solution I get `(gdb) p (*self.get()) You can't do that without a process to debug.` – Rostfrei Dec 18 '18 at 14:56

vacing · Answer 3 · 2020-05-22T01:50:31.827

Answer first:

p *frame._M_ptr # frame is the shared_ptr's name

I tried p (*frame.get()), but it didn't work(frame is my shared_ptr name)

(gdb) p frame
$4 = std::shared_ptr (count 2, weak 0) 0x2ea3080
(gdb) p (*frame.get())
Cannot evaluate function -- may be inlined

then I tried to get what's in this shared_ptr, then I found this

(gdb) p frame.
_M_get_deleter  __shared_ptr    operator*       reset           unique          ~shared_ptr     
_M_ptr          get             operator->      shared_ptr      use_count       
_M_refcount     operator bool   operator=       swap            ~__shared_ptr

I used it's _M_ptr field, it worked.

(gdb) p *frame._M_ptr 
$5 = {
...
}

I used std::shared_ptr, and gdb 7.6.

How to access target of std::tr1::shared_ptr in GDB

3 Answers3

Linked