I am trying to get a better understanding of std::enable_if
in C++11 and have been trying to write a minimal example: a class A
with a member function void foo()
that has different implementations based on the type T
from the class template.
The below code gives the desired result, but I am not understanding it fully yet. Why does version V2
work, but not V1
? Why is the "redundant" type U
required?
#include <iostream>
#include <type_traits>
template <typename T>
class A {
public:
A(T x) : a_(x) {}
// Enable this function if T == int
/* V1 */ // template < typename std::enable_if<std::is_same<T,int>::value,int>::type = 0>
/* V2 */ template <typename U=T, typename std::enable_if<std::is_same<U,int>::value,int>::type = 0>
void foo() { std::cout << "\nINT: " << a_ << "\n"; }
// Enable this function if T == double
template <typename U=T, typename std::enable_if<std::is_same<U,double>::value,int>::type = 0>
void foo() { std::cout << "\nDOUBLE: " << a_ << "\n"; }
private:
T a_;
};
int main() {
A<int> aInt(1); aInt.foo();
A<double> aDouble(3.14); aDouble.foo();
return 0;
}
Is there a better way to achieve the desired result, i.e. for having different implementations of a void foo()
function based on a class template parameter?
I know this wont fully answer your question, but it might give you some more ideas and understanding of how you can use std::enable_if
.
You could replace your foo member functions with the following and have identical functionality:
template<typename U=T> typename std::enable_if<std::is_same<U,int>::value>::type
foo(){ /* enabled when T is type int */ }
template<typename U=T> typename std::enable_if<std::is_same<U,double>::value>::type
foo(){ /* enabled when T is type double */ }
A while back I gained a pretty good understanding of how enable_if works, but sadly I have forgotten most of its intricacies and just remember the more practical ways to use it.
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