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

Works for me:

g++ -g test.cpp -o test
gdb test
(gdb) break main
(gdb) r


Breakpoint 1, main () at test.cpp:22
22      A::B *p = new A::B(100);
(gdb) n
24      p->print();
(gdb) n
m_a is 100
26      int *q = 0;
(gdb) p p
$1 = (A::B *) 0x602010
(gdb) p (A::B *) 0x602010
$2 = (A::B *) 0x602010
(gdb) p *((A::B *) 0x602010)
$3 = {m_a = 100}

It works for me. What are you using (gcc version, OS, compilation flags?)