In C++, how to get the address in which the value of a class instance's field is stored?

Question

class A {
   public: int i;
};

A *a = new A();

How to get the address of a->i? I tried &a->i and also &(a->i) but those generate compile time errors:

"left of '.i' must have class/struct/union type"

Answer 1

You have not provided the same code you tried to compile. Always copy and paste. The tells in your code are that you don't have a syntactically correct class declaration or variable declaration, and that your error message talks about ".i" when you've claimed you've only used a->i. Here's working code:

#include <stdio.h>
class A {
    public:
        int i;
};

int main() {
    A* a = new A();
    int* i = &a->i;
    printf("a: %p\na->i: %p\n", a, i);
    return 0;
}

Ultimately, the syntax you say you tried for getting the address of the member was correct. The syntax the error message says you tried was a.i. That doesn't work, and for the reason the error message gave. The variable a is not a class, struct, or union type. Rather, it's a pointer to one of those types. You need to dereference the pointer to get at the member.

When I run it, I get this:

$ ./a.out
a: 40748
a->i: 40748

The addresses are the same because A is a simple class, so this output is to be expected. The first member is frequently placed at the very start of a class's memory. Add a second member variable to the class and get its address; you should see different values then.

Answer 2

In c++:

class A {
    public:         // note colon after public
        int i;
};                  // note semicolon after bracket

A *a = new A();     // note *a not a*

to print:

cout << ios::hex << &a->i << endl;

For me this seems to work ok.

Answer 3

&a->i should work. In your case since the class just has one public integer both the address of a and i will be the same.

Answer 4

You're on the right track, but from the compile error you mention, it sounds like you were typing "&a.i"

class A 
{
   public: 
       int i;
};

...

A *a = new A();
int *i = &a->i;

Answer 5

This code appears to compile?

class A {
   public:
      int i;
};
int main() {
   A *a = new A();
   int *x = &a->i;
   return 0;
}

Answer 6

The following works for me using g++

class A
{
public:
    int m_i;
};

int
main()
{
    A* a = new A();
    int* i_ptr = &(a->m_i);

    return 0;
}

I am guessing you mean A* a = ... and not A a* = ...