Why a call to a consteval from object constructor is not constant expression?

Question

I have the consteval function shown below:

template <std::size_t text_length>
consteval std::size_t text_id(const char(&a_text)[text_length])
{
    std::size_t result{text_length};
    for (const auto &c : a_text)
    {
        result ^= c;
        result <<= ((c % 7u) + 1u);
    }
    return result;
}

It works as expected on the places where a compile-time value is expected:

template <auto x>
auto v = x;

int main(int argc, char **argv)
{
    constexpr auto id = text_id("test");

    switch (argc)
    {
        // No problem
        case text_id("test"):
            std::cout << v<text_id("test")>; // No problem either
            break;
    }

    return 0;
}

But it doesn't compile if I put it on an object constructor:

struct S
{
    template <std::size_t size>
    constexpr S(const char (&text)[size]) :
        id{text_id(text)}
    {}

    std::size_t id;
};

int main()
{
/*
error: 'text' is not a constant expression
        id{text_id(text)}
           ~~~~~~~^~~~~~
*/
    constexpr S s("test");
    static_assert(text_id("test") == s.id);

    return 0;
}

Is there a way to make it work?

you simply cannot do it, call a consteval function with potentially non-consteval value make little sense, if you want, you can declare the constructor as consteval. or the function constexpr. — apple apple, Mar 18 '22 at 18:36
Hold on! I didn't know that `S::S` can be marked as `consteval`! that totally works! :) — PaperBirdMaster, Mar 18 '22 at 18:39
`constexpr` allows evaluation at run time while `consteval` requires compile time evaluation. So `constexpr` needs to generate run time code for the function with an unknown `text` even if you never call it. The call could be in another compilation unit. — Goswin von Brederlow, Mar 19 '22 at 01:05

apple apple · Answer 1 · 2022-03-18T19:03:03.893

1

You cannot call consteval function with constexpr function's parameter, since it may be non-constant.

consteval int bar(int){
    return 1;
}

constexpr int foo(int s){
    return bar(s); // error
}

You can change S::S to consteval or change text_id to constexpr.

edited Mar 18 '22 at 19:03

answered Mar 18 '22 at 18:54

apple apple

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btw, call `constexpr` function with `consteval` function's parameter is also valid. (so `consteval S::S` and `constexpr text_id` also works) – apple apple Mar 18 '22 at 19:10

Why a call to a consteval from object constructor is not constant expression?

1 Answers1