Passing a point in to a function

Question

In this program I've created two pointers (a,b) that points to the memory address of x and y. In the function I've created its supposed to swap the memory address of a and b(So b=a and a=b). When I compile it gives me a error (invalid conversion from 'int' to 'int*') What does that mean? I'm passing a pointer to the function or is it read it as a regular int?

#include <iostream>
using std::cin;
using std::cout;
using std::endl;
void pointer(int* x,int* y)// Swaps the memory address to a,b
{
    int *c;
    *c = *x;
    *x = *y;
    *y = *c;
}

int main()
{
    int x,y;
    int* a = &x;
    int* b = &y;
    cout<< "Adress of a: "<<a<<" Adress of b: "<<b<<endl; // Display both memory address

    pointer(*a,*b);
    cout<< "Adress of a: "<<a<<" Adress of b: "<<b<<endl; // Displays the swap of memory address  

    return 0;
}

error message:

C++.cpp: In function 'int main()':
C++.cpp:20:16: error: invalid conversion from 'int' to 'int*' [-fpermissive]
C++.cpp:6:6: error: initializing argument 1 of 'void pointer(int*, int*)' [-fpermissive]
C++.cpp:20:16: error: invalid conversion from 'int' to 'int*' [-fpermissive]
C++.cpp:6:6: error: initializing argument 2 of 'void pointer(int*, int*)' [-fpermissive]

Answer 1

*a and *b are of type int, while a and b are both of type int*. Your function takes two int*'s, so all you have to do is change

pointer(*a,*b);

to

pointer(a,b);