How to initialise and return an std::array within a template function?

Question

I'd like a template function that returns some initialisation.

I did the following:

#include<iostream>
#include<array>

template <typename T>
inline static constexpr T  SetValues()
{
    if(std::is_integral<T>::value)
    {
      std::cout<<"int\n";
      return T{1};
    }
    else
    {
      std::cout<<"array\n";
      T arr = {1,2,3,4};
      return arr;
    }
}


int main()
{
 int a = 6;
 std::array<int, 4> arr = {2,2,2,2};
 a = SetValues<decltype(a)>();
 arr = SetValues<decltype(arr)>();
 return 0;
}

which correctly initialises the int but in case of the array I get the error

scalar object ‘arr’ requires one element in initializer

How should I initialise and return the array?

paolo · Accepted Answer · 2022-05-12 08:47:13Z

2

The problem is that, when T = int, you're trying to trying to initialize int arr with initializer list {1, 2, 3, 4} in the else branch. Even if you don't execute the else branch, the compiler still has to compile it and int arr = {1, 2, 3, 4} is invalid syntax.

On C++17 (and higher), you can get away with if constexpr:

template <typename T>
inline static constexpr T SetValues() {
    if constexpr (std::is_integral<T>::value) {
        std::cout << "int\n";
        return T{1};
    } else {
        std::cout << "array\n";
        T arr = {1, 2, 3, 4};
        return arr;
    }
}

With if constexpr, the compiler won't compile the else branch when T is an integral type (e.g. int) and will not compile the if branch when T is not an integral type.

edited May 12, 2022 at 8:47

answered May 12, 2022 at 8:40

paolo

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