if constexpr fails when passing value as parameter not when passed as template

Asked May 05 '20 at 18:51

Active May 05 '20 at 18:51

Viewed 36 times

I have some code looking like this

#include <array>

constexpr std::size_t N = 16;

template<typename Array>
constexpr void takeparam(Array& data, size_t I) {
    if  (I < N - 2) { // <- fails if I add constexpr
       data[I] = data[I + 2];
    }
}

template<size_t I, typename Array>
constexpr void taketemplate(Array& data) {
    if  constexpr (I < N - 2) { // <- here constexpr is OK
       data[I] = data[I + 2];
    }
}

int main() {
    std::array<int, 16> data = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 };

    takeparam(data, 3); 
    taketemplate<3>(data); 
}

I was expecting to be able to use if constexpr in the takeparam function, but I cannot get it work. What am I doing wrong?

asked May 05 '20 at 18:51

OznOg

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Inside of a `constexpr` function, its parameters are never considered to be `constexpr`. – HolyBlackCat May 05 '20 at 18:52
I don't understand what the problem is here? This compiles for me. – Martin York May 05 '20 at 18:55
`if constexpr` is evaluated at compile-time, but `I < N - 2` cannot be evaluated at compile-time if `I` is a run-time parameter rather than a compile-time constant. – Remy Lebeau May 05 '20 at 19:19

if constexpr fails when passing value as parameter not when passed as template

0 Answers0