用于特定类型推导的功能模板专门化

Question

class base
{
};

class derived
{
};

template<class T> void foo() {}


int main()
{
    foo<int>();
    foo<derived>();
}

I want to specialize foo for T = derivation of base . Is this possible or do I need to specialize for base itself?

Answer 1

You can combine boost::enable_if and boost::is_base_of, as is documented in the manual for boost::enable_if .

template <class T>
T foo(typename enable_if<boost::is_base_of<base,T> >::type* dummy = 0); 

Answer 2

One option would be to use boost type traits (or if feeling brave, look at the source of that library).

At compile time you can detect if a type inherits from another type and so choose an appropriate implementation.