Overloading templated functions

Question

I have a doubt about if that approach is possible, lets say you want two ways of calling a function at same time, one is returning an object and the other return by reference in parameter:

// ... template <class T> void func(Foo<T>& f, const T n) { f.a = Something(f.a + n); f.b = Something(f.b + n); } template <class T> Foo<T> func(const Foo<T>& f, const T n) { return Foo<T>( Something(f.a + n), Something(f.b + n) ); } // ... // main Foo<int> foo(1, 1); func(foo, 2); Foo<int> foo2 = func(foo, 2);

The const word in first parameter affect the signature of method?

I don't think you can do it taht way

Sam I am says Reinstate Monica
– Sam I am says Reinstate Monica

2012-02-28 15:06:07 +00:00
Commented Feb 28, 2012 at 15:06 — Sam I am says Reinstate Monica
– Sam I am says Reinstate Monica, Commented Feb 28, 2012 at 15:06

Luchian Grigore · Accepted Answer · 2012-02-28 15:13:53Z

Yes it does. const affects the signature, the return type does not.

1.3.11 signature

the information about a function that participates in overload resolution (13.3): its parameter-type-list (8.3.5) and, if the function is a class member, the cv-qualifiers (if any) on the function itself and the class in which the member function is declared. [...]

const is part of hte parameter-type-list, so it does determine an overload.

Mike Seymour · Accepted Answer · 2012-02-28 15:14:36Z

Yes, a reference to const is a separate type to a reference to non-const, so the two functions with these arguments are separate overloads.

However, this will not work:

Foo<int> foo2 = func(foo, 2);

Since foo is not const, this will select the non-const overload, which has no return value; so the assignment will fail. You would need to explicitly choose the const version:

Foo<int> foo2 = func(static_cast<const Foo<T>&>(foo), 2);

@JoachimPileborg: Because const_cast is a more dangerous cast than is needed; in this case, it could remove a volatile qualifier. (Although static_cast could perform an unsafe type conversion; unfortunately, there's no explicit cast that only allows standard conversions).

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