Question

I need to clear a basic concept. This code works fine. Can somebody explain me that if the function calDouble already returning the address (reference) of int why I need to use & operator further in main int *j = &calDouble(i); to get the address (reference) of int? Thanks.

int& calDouble(int x)
{
    x = x*2;
    return x;

}

int main(int argc, char *argv[])
{
    int i = 99;
    int *j = &calDouble(i);

    system("PAUSE");
    return EXIT_SUCCESS;
}

Answer 1

int& calDouble(int x) doesn’t return an address, but a reference to an int.

You need to take the address of the reference to be able to assign it to a pointer.

Note however that your code invokes undefined behavior. Because you pass the parameter by value, a copy of it is created inside the function. So you return a local variable by reference, which is not legal.

I think your confusion comes from &. This can be used in two ways:

• applied to a variable &x, it takes its address
• when in a declaration int& x, it defines a reference

A reference is just an alias, a different name for a variable.

int x = 0;
int& y = x;

Now, x and y refer to the same variable.

int* z = &x;

takes the address of x.