Re: enable_if and pure virtual functions

=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <>
Thu, 12 Jan 2012 15:08:44 -0800 (PST)
Am 12.01.2012 20:48, schrieb kelvSYC:

I'm kind of confused on how enable_if and virtual functions work
together. Suppose you have this:

(using your favourite implementation of enable_if here)

template<class T>
struct DerivedProperty;

template<class T>
class AbstractBase { // CRTP, as you can't mix "virtual" and
"function template", it would appear... or can you?

Yes, you cannot declare a function template that is a virtual function.

    virtual enable_if<DerivedProperty<T>, int>::type doSomething() =

This is an absolutely useless application of enable_if - and this has nothing to do with virtual functions. In your example the declaration of doSomething() will be instantiated once AbstractBase<U> will be instantiated for some type U making the instantiation of AbstractBase<U> potentially ill-formed (depending on U). The only reason for such a construction would be to act as some convoluted form of static assertion that would better be written as

template<class T>
class AbstractBase {
 static_assert(DerivedProperty<T>::value, "DerivedProperty not satisfied");
 virtual int doSomething() = 0;

instead. If you want to declare a virtual function depending on some constraints, it is better to do that via a base-class switch, e.g.

template<class T, bool = DerivedProperty<T>::value>
class AbstractBase {
 virtual int doSomething() = 0;

template<class T>
class AbstractBase<T, false> {};

Also, would something similar apply if the enable_if was instead a
parameter instead of the return type?

This would not change anything, because in either situation the declaration of the function could be ill-formed.

HTH & Greetings from Bremen,

Daniel Kr?gler

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