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?

Parameters for which no argument has been provided (at the call site) and which have default arguments are ignored for overload resolution [over.match.viable]/2. — dyp
– dyp, Commented Nov 13, 2013 at 22:23

Community · Accepted Answer · 2017-05-23 11:58:31Z

2

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.

edited May 23, 2017 at 11:58

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answered Nov 13, 2013 at 22:21

Sam

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3 Comments

dyp Over a year ago

ADL? You mean overload resolution? This has nothing to do with name lookup.

Sam Over a year ago

@DyP: Thank you. As far as I know, it depends on the actual context, whether ADL is used additionally to a name lookup. It may be applied to gather other candidates for overload resolution. Mentioning ADL in conjunction with the OP's code snippet was misleading, so I removed that part.

dyp Over a year ago

"whether ADL is used additionally to a name lookup" ADL is a form of name lookup; it's always used when the function name is unqualified (qualified: my_class_or_namespace::f; unqualified: f). Btw you included the link to cppreference twice ;)

pointhi · Accepted Answer · 2013-11-13 22:20:34Z

1

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);

edited Nov 13, 2013 at 22:20

answered Nov 13, 2013 at 22:15

pointhi

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