24

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;

}

Community
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Gary Hu
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2 Answers2

21

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;

R Yoda
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Matt
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16

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?)

sehe
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  • It's wired! why I found that: (gdb) p p $1 = (A::B *) 0x1878f010 (gdb) p (A::B *) 0x1878f010 A syntax error in expression, near `) 0x1878f010'. (gdb) p *((A::B *) 0x1878f010) A syntax error in expression, near `) 0x1878f010)'. – Gary Hu Sep 12 '11 at 13:03
  • it's wired? I hope that is a good thing. The rest of the comment didnot make sense to me, unfortunately – sehe Sep 12 '11 at 13:07
  • I use the same compilation flags with you. My os is CentOS 5 x64. GDB version is: GNU gdb Fedora (6.8-27.el5) Copyright (C) 2008 Free Software Foundation, Inc. License GPLv3+: GNU GPL version 3 or later This is free software: you are free to change and redistribute it. There is NO WARRANTY, to the extent permitted by law. Type "show copying" and "show warranty" for details. This GDB was configured as "x86_64-redhat-linux-gnu"... – Gary Hu Sep 12 '11 at 13:08
  • you could use another variable than p... – sop Sep 29 '16 at 11:37
  • @sop Yes. Why do you mention it? – sehe Sep 29 '16 at 11:39
  • it's a little confusing :P – sop Sep 29 '16 at 11:58