How to get std::array's size at compile time

[kshegunov]

Hello,
I have a function which returns a std::array and I want to have some logic based on the size of the array at compile time. Is there a way to get the size of the array off the return value of my function. Code looks something like this:

auto someFunction()
{
    return std::array<double, 14>{ 0.1, ..., ...};
}

And I want to plug it into a if constexpr to branch out a specific case. Something akin to this:

constexpr decltype(someFunction().size()) size = ...;
if constexpr (size > 10)  {
   // Do special stuff
}

Is this possible at all?

[VRonin]

I's very easy, you just use std::tuple_size. The naming is misleading but it does its job

#include <iostream>
#include <array>
#include <type_traits>

auto someFunction(){
    return std::array<double,3>{1.1,2.2,3.3};
}

int main()
{
    if constexpr (std::tuple_size<std::result_of<decltype(someFunction)&()>::type>::value > 1)  
        std::cout<<"More than 1";
    return 0;
}

[mrjj]

Hi
Or like this which also seems to work

#include <array>

void Nurse();
void Nuke();

constexpr auto someFunction() {
    return std::array<double, 14>{ 0.1};
}

void test() {
constexpr  decltype( someFunction().size()  ) size = someFunction().size();
if constexpr (size > 10)  {
   Nuke();
} else {
   Nurse();
  }
}

see here
https://godbolt.org/z/Ktt36G

But @VRonin version seems nicer :) \o/

[kshegunov]

Thanks for the input guys, much appreciated!