Question

#include<iostream>

using namespace std;

void passPointer(int *pointer)
{
    cout << *pointer;
}

int main()
{
    int *iNum = new int(25);

    passPointer(iNum);

    return 0;
}

Can someone explain to me why when I use the passPointer() function in main, it has to be passPointer(iNum) but not passPointer(*iNum)? Is it because I am dereferencing it at the parameter if I use *? Please explain as detailed as you can as I am a bit confused.

Thanks guys.

Answer 1

I am very sympathetic to these sorts of questions, because this is one of the only things that I had trouble with when learning C++.

The basic problem is that in the syntax of C++, the * and & characters are used for many different things with similar but subtly different meanings.

In your case, you are considering using * in four different places.

In the first place: int *iNum = new int(25);, the * is sitting in a declaration. This means that is is a type annotation saying that iNum is a pointer.

In the second place: passPointer(*iNum);, the * is sitting in an expression. This means that it is the dereference operator, which means: “get the value pointed to by iNum“. In this case the value pointed to by iNum is an int. As you will see later, passPointer is declared to take an argument of type pointer to int, so you cannot pass a plain int as an argument to passPointer. You should instead just pass iNum (as iNum is a pointer to int).

In the third place: void passPointer(int *pointer), the * is again sitting in a declaration. This means that it has the same meaning as in the first place – it says that pointer is a pointer (to int).

In the fourth place: cout << *pointer;, the * is again sitting in an expression. This means that, as in the second case, it is saying “dereference pointer and get the value that pointer is storing the address of”.