CODEWITHSUNDEEP
c++type-traits
Blood-HaZaRd
Blood-HaZaRd

Reputation: 2124

Can't deduct template parameter type using std::conditional

I would like to call a function template like below 

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

struct size1 { size1() {std::cout << "Caling 1 \n";}};
struct size2 { size2() {std::cout << "Caling 2 \n";}};

template <typename T, typename std::conditional_t<sizeof(T) == 4, size1, size2> U>
void afficher(T a)
{
    std::cout << typeid(U).name();
}


int main(int argc, char *argv[])
{
    afficher(10); //Error can't deduct U
}

I think that here I have a non-deductable context, how could I correct it and

is it ok to user std::condittional here or use std::enable_if ?

Thank you.

Upvotes: 1

Views: 78

Answers (1)

jrok
jrok

Reputation: 55395

You've got a syntax problem, nothing else:

template <typename T, typename U = std::conditional_t<sizeof(T) == 4, size1, size2>>
void afficher(T a)         //  ^^^^
{
    std::cout << typeid(U).name();
}

As noted by Jarod42 in the comments, this allows users to bypass your intent and do whatever with the second argument. You could use a typedef instead:

template <typename T>
void afficher(T a)
{
    using U = std::conditional_t<sizeof(T) == 4, size1, size2>>;
    std::cout << typeid(U).name();
}

Upvotes: 5

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