How to print a c++ object members using GDB from an address if the object's class type is like A::B [duplicate]

Question

From this link gdb interpret memory address as an object we know that, if an object of class type A is at a specific address such as 0x6cf010, then we can use:

(gdb) p *(A *) 0x6cf010 

to print the member elements of this object.

However, this seems doesn't work when c++ namespace is involved. That is, if the object of class type A::B, then all the following trying doesn't work:

(gdb) p *(A::B *) 0x6cf010
(gdb) p *((A::B *) 0x6cf010)

So, who knows how to print the object elements under this conditions?

---

We can use the following deliberate core code to try to print the members of p from the address (we can use "info locals" to show the address).

#include <stdio.h>

namespace A
{
    class B
    {
    public:
        B(int a) : m_a(a) {}

        void print()
        {
            printf("m_a is %d\n", m_a);
        }

    private:
        int  m_a;
    };
}

int main()
{
    A::B *p = new A::B(100);

    p->print();

    int *q = 0;

    // Generating a core here
    *q = 0;
    return 0;

}

Answer 1

I know that this is labeled as answered, but I was able to reproduce this problem using gdb on OS X (GNU gdb 6.3.50-20050815 (Apple version gdb-1820) (Sat Jun 16 02:40:11 UTC 2012)) and the works-for-me solution didn't answer it for me.

Turns out there was another question on SO that did have an answer which worked, so I think it's worth pulling into this quesiton:

Why gdb casting is not working?

The short answer is that you may have to single-quote your namespaced variables:

(gdb) p ('MyScope::MyClass'*) ptr;