how to pass a value to its parameter (default parameters)

Question

I'm really confused because I can't describe my question well, but I'm sure that many of you will understand me.

#include <iostream>

using namespace std;

void display(int n = 1, char c = '*');

int main()
{
    display();
    display(5);
    display('$');
    return 0;
}

void display(int n, char c)
{
    for (int i = 1; i <= n; i++)
        cout << c;
    cout << endl;
}

At display('$'), I want to pass this char to its parameter c and use n with its default value, which is 1. Can anyone tell me how to do this properly?

Answer 1

You can't as is. When you call a function the parameters get matched from left to right in order. That means display('$'); will give '$' to n instead of c.

What you can do though, at least in this case, is overload the function to do what you want. With

void display(int n, char c);
void display(int n);
void display(char c);
void display();

You can have the void, int and char overload call the main function and hide the fact that your filling in the "blanks". That looks like

void display(int n)
{
    display(n, '*');
}

void display(char c)
{
    display(1, c);
}

void display()
{
    display(1, '*');
}

The down side here is your repeating the default values. That makes this brittle as a change of the default values requires you to change it in multiple places.