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.

Collazo

Which compiler and stdlib are you using? Also which version of GDB? – Parenthood

@Parenthood sorry! Updated my question. – Collazo

have you tried (gdb) p sharedPtr.get()? – Tomsk

@Raydel Miranda That worked! Thanks very much! I'll add answer in a moment. – Collazo

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? – Collazo

unique_ptr: #22799101 – Appertain

Answer 1

Try with

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

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

Tomsk

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. – Collazo

If I use this solution I get (gdb) p (*self.get()) You can't do that without a process to debug. – Watchtower

Answer 2

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.

Imperfective

You meant ptr.get() right? I think the operator -> is not what you want use here. – Tomsk

ptr->get() means *ptr.get(). – Imperfective

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()). – Tomsk

Answer 3

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.

Dodeca