Constantly inlining a function with a consteval call

Question

I have a consteval crc32 function that works just fine on compile time. I want this function to be used in another constant inline function.

Here is a code example for a better understanding:

unsigned int consteval strsum(const std::string str) {
    auto n = 0;

    for (int i = 0; i < str.size(); i++) n += str.at(i);

    return n;
}

void _printHash(const unsigned int hash) {
    printf("%d", hash);
}

__forceinline void printHash(const std::string str) {
    _printHash(strsum(str));
}

int main(int argc, char* argv[]) {
    printHash("abc");
}

I want this code to compile to just:

int main(int argc, char* argv[]) {
    _printHash(/* hash constant */);
}

Instead, i get a C7595 error: A call to a consteval function is not a constant expression.

Is there any way to acheive wanted behavior on MSVC?

Answer 1

First thing first, you cannot have a consteval function with an std::string parameter because it's not a literal type. Fortunately std::string_view is infinitely better in this application, and you can have

consteval unsigned int strsum(const std::string_view str) {
    auto n = 0;
    for (auto s : str) n += s;
    return n;
}

Sadly, a parameter of a non-consteval function is never a constant expression, and a consteval function cannot call a regular function. So even though you can have strsum, you cannot call it the way you want.

You could call _printHash(strsum("abc")) instead of printHash("abc"), but if this syntax is not acceptable to you, then I'm afraid you don't have a lot of options, other than accepting that the resulting code will loop over a constant string.

A compiler that is not MSVC will do the right thing without needing any constexpr or consteval (or inline for that matter), so the entire thing is compiled to something equivalent to printf("%d", 294) exactly as desired. But MSVC is just not very good at this kind of optimisations.