Template type deduction fails?

Question

Consider the following example.

#include <type_traits>
#include <iostream>
using namespace std;

template <typename T_> 
using Integral = typename std::enable_if<std::is_integral<T_>::value,T_>::type;
template <typename T_> 
using NotIntegral = typename std::enable_if<!std::is_integral<T_>::value, T_>::type;

template <typename T_>
void printIt(const Integral<T_> &value) { cout << "Integral == " << value << endl; }

template <typename T_>
void printIt(const NotIntegral<T_> &value) { cout << "Non Integral == " << value << endl; }

template <typename T_>
void foo(const T_ &value) { printIt<T_>(value); }

int main(int argc, char** argv)
{
    printIt<int>(66);   //Must explicitly provide argument type.
    //printIt(33);        //Compiler error. No overloaded function....????
    foo(29.); 

    return 0;
}

Why do I need to explicitly set the type of template parameter? Should the compiler figure out it's an int type argument?

Answer 1

Why do I need to explicitly set the type of template parameter?

Because these are non-deduced contexts.

Imagine specializing std::enable_if<std::is_integral<T_>::value,T_> so that ::type evaluates to something else. The compiler cannot know the mapping from typename something<T>::type to T.

You can achieve your desired result by placing std::enable_if as part of the return type, so that non-matching overloads are SFINAEd out:

template <typename T>
auto printIt(T x) -> std::enable_if_t<std::is_integral_v<T>, void> { /*...*/ }

template <typename T>
auto printIt(T x) -> std::enable_if_t<!std::is_integral_v<T>, void> { /*...*/ }

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