Using the default value for a leading argument while specifying the others

Question

I am wondering how I can skip overriding a parameter in a C++ function. For example, looking at the function below, if output has 1 parameter, you can just call it without sending any parameters, like output();

That will output 5 since xor has a default value of 5. However, if I want to override "vor", and leave xor to its default value, how do I do that?

output(NULL, 20);

Above didn't work, it just initalized xor to 0.

void output(int xor = 5, int vor = 15) { cout << xor << " " << vor << endl; } int main() { output(10, 20); }

The Named Parameter idiom might help: isocpp.org/wiki/faq/ctors#named-parameter-idiom — Christian Hackl
– Christian Hackl, Commented Aug 23, 2014 at 16:36

Vlad from Moscow · Accepted Answer · 2014-08-23 15:42:52Z

If you want to override the second default argument then you have to specify the first argument.

Possible calls of the function are following

output(); // corresponds to output( 5, 15 ); output( x ); // corresponds to output( x, 15 ); output( x, y ); // corresponds to output( x, y );

where x and y are some arbitrary arguments.

wangjunwww · Accepted Answer · 2014-08-23 15:48:20Z

To purely realize what you want to achieve, you can switch the parameter order in output() like this:

void output(int vor = 15, int xor = 5) { cout << xor << " " << vor << endl; }

Then you can call output(20) to override vor while keeping xor to its default value.

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