Function overloading resolution

Question

void f();
void f(int);
void f(int, int);
void f(double, double = 3.14);
f(5.6);  // calls void f(double, double) and not f(int) or f(), for that matter. Why?

I read that the compiler checks the number of parameters before checking the type of parameters. Then why aren't all the functions having different number of parameters getting eliminated?

Answer 1

It does call void f(double, double = 3.14);, because of the default value for the second argument; one double provided, one required -> match. Otherwise, void f(int); would be selected. So its the number of mandatory parameters that matters.

Further info:

• Overload resolution: http://en.cppreference.com/w/cpp/language/overload_resolution
• Implicit conversion rules: Implicit type conversion rules in C++ operators.

Answer 2

You have defined the second value from the function:

void f(double, double = 3.14);

So the call

f(5.6);

is like

f(5.6, 3.14);

use explicit type converting, to call the other function, like:

f((int)5.6);