CODEWITHSUNDEEP
c++templates
user109078
user109078

Reputation: 906

Any way to fix this type deduction?

Type deduction fails for the following case. It compiles if I specify the template argument to someFunc. I definitely see how this is a weird case but it would be nice if I could get it to work. Is there an alternate way to formulate it that would compile without providing the template parameter? A C++17 solution is fine.

#include <type_traits>

template<typename T>
using choose_arg_type = typename std::conditional<std::is_fundamental<T>::value,T,const T &>::type;

template<typename T>
T someFunc(choose_arg_type<T> arg)
{
    return arg + arg;
}

int main()
{
    auto result = someFunc(0.0);

    return 0;
}

Upvotes: 4

Views: 134

Answers (2)

SergeyA
SergeyA

Reputation: 62613

No, choose_arg_type<T> is in non-deduced context. There are quite a few duplicates already.

However, you can have two overloads, and SFINAE-enable them based on the type.

Upvotes: 0

NathanOliver
NathanOliver

Reputation: 181007

In

template<typename T>
T someFunc(choose_arg_type<T> arg)

T is a dependent type. Because of that no type deduction will happen here. You can work around this issue by using SFINAE and introducing a set of overloads for when the type is fundamental or not. That would look like

template<typename T, std::enable_if_t<std::is_fundamental_v<T>, bool> = true>
T someFunc(T arg)
{
    return arg + arg;
}

template<typename T, std::enable_if_t<!std::is_fundamental_v<T>, bool> = true>
T someFunc(const T& arg)
{
    return arg + arg;
}

Upvotes: 5

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