Cannot use static_assert inside a constexpr function inside a constexpr object

Question

I want to check at compile time if some string is in a list of strings. So if I use static_assert in the main, this works. But if I want to use static_assert inside a constexpr function it gives me compile errors:

Any idea on why this compile error and how to make this compile time check work?

#include <array>
#include <string_view>

class ClosedList : public std::array< std::string_view, 3>
{
public:    
    constexpr bool hasValue( std::string_view  value) const
    {
        for (auto& e : *this) {
            if (value == e) {
                return true;
            }
        }
        return false;
    }

    constexpr std::string_view value(std::string_view value) const
    {
        static_assert( hasValue( value ), "value not in set");
        return value;

    }
};

int main()
{
    constexpr ClosedList myList = { "red", "green", "blue" };    
    static_assert(myList.hasValue( "red" ), "value not in set" );
    auto value = myList.value("red"); // compile error
}

Inside of a constexpr functions, the parameters and this don't count as constexpr. The function doesn't have to be called on a compile-time constant. — HolyBlackCat
– HolyBlackCat, Commented Nov 6, 2020 at 21:57
This is a question&answer forum. In the spirit of this forum, please ask a question. The behavior of your compiler seems correct. Please take a tour and read How to Ask. — KamilCuk
– KamilCuk, Commented Nov 6, 2020 at 22:07
@HolyBlackCat hasValue() uses *this and still is constexpr, so I don't see why value() cannot use this function and still being constexpr. — Martijn Breen
– Martijn Breen, Commented Nov 6, 2020 at 22:08
@MartijnBreen: You seem to be misunderstanding what the constexpr modifier on a function does. It causes the compiler to attempt compile-time evaluation IN A CONSTANT CONTEXT. The compiler can always do compile-time evaluation under the as-if rule, but without constexpr the result cannot be used in a constant context. But as HolyBlackCat said, functions marked constexpr can also be used outside constant context, so they can't guarantee that the static_assert parameter is a constant expression. — Ben Voigt
– Ben Voigt, Commented Nov 6, 2020 at 22:18

parktomatomi · Accepted Answer · 2020-11-07 15:39:18Z

0

If you need to assert something in a constexpr method, just throw. Really. As long as the throw expression isn't actually evaluated in a constant context, it's explicitly allowed to compile and be used in constexpr functions and methods:

if (!hasValue(value)) {
    throw std::runtime_error("value not in set");
}

Then it works like you want:

constexpr ClosedList myList { "a", "b", "c" };
constexpr std::string_view b = myList.value("b"); // ok
constexpr std::string_view d = myList.value("d"); // error, not a constant expression

The compile error even points to the line with the throw so you can see the error, kind of.

Demo: https://godbolt.org/z/da36dz

answered Nov 7, 2020 at 15:39

parktomatomi

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