How can I move or assign one vector into another depending of their type using if constexpr?

Question

I wanted to move or assign a std::vector<Scalar> into a std::vector<float> depending on the type Scalar.

If Scalar is float then move, else copy.

I tried this code

#include <vector>

using Scalar = double;

int main()
{
    std::vector<Scalar> x(10);
    std::vector<float> y(10);

    if constexpr(std::is_same<Scalar, float>::value)
        y = std::move(x); // line 11 (does not compile)
    else
        y.assign(x.begin(), x.end());

    return 0;
}

but I get

main.cpp:11:24: error: no match for ‘operator=’ (operand types are ‘std::vector<float>’ and ‘std::remove_reference<std::vector<double>&>::type’ {aka ‘std::vector<double>’})
   11 |         y = std::move(x);
      |                        ^

I thought that since std::is_same<Scalar, float>::value is evaluated to false at compile-time, then the line below would not be compiled.

I compiled it using g++ -std=c++17 main.cpp -o exec.

How can I move/assign x into y depending on the Scalar type?

Answer 1

Because main() isn't a template, both sides of the if constexpr must be valid. To use if constexpr this way, it needs to be in a template.

Thankfully, a good template makes this logic more reusable:

#include <vector>

template<typename T, typename U>
void move_or_copy(std::vector<T>& dest, std::vector<U>&& src)
{
    if constexpr(std::is_same_v<T,U>)
        dest = std::move(src);
    else
        dest.assign(src.begin(), src.end());
}



using Scalar = double;

int main()
{
    std::vector<Scalar> x(10);
    std::vector<float> y(10);

    move_or_copy(y, std::move(x));
}

We could make the template function better constrained (allow any collection as source), but this should at least provide the correct direction in which to depart.