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c++ constexpr consteval std-source-location

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August 5, 2025 Score: 6 Rep: 229,739 Quality: Expert Completeness: 80%

Function parameters are not constant expression, even in consteval function.

You have to pass argument as template.

Since C++20, you might have structural types as template parameter, so

constexpr std::sizet constexprstrlen(const char s)
{
    std::size_t i = 0;
    while (s[i]) ++i;
    return i;
}
template  // Some hard coded limit for non literal string
struct LiteralString
{
    consteval LiteralString(const char s) { std::copy(s, s + constexprstrlen(s), &data[0]); }
    consteval LiteralString(const char (&s)[N]) { std::copy(s, s + N, &data[0]); }
    static constexpr std::sizet size = N;
    std::array data{};
};

Then you might do something like:

template 
void foo() {
    constexpr std::sizet len = constexprstrlen(S.data.data());
    // ...
}
int main() {
    foo();   // empty string for clang/msvc :(
    foo(); // empty string for clang/msvc :(
}

Unfortunately, only gcc feeds current sourcelocation in template parameter Demo.

MACRO might be a workaround Demo

#define FOO() foo()
int main() {
    foo();   // empty string for clang/msvc :(
    foo(); // empty string for clang/msvc :(
    // "int main()" or "int _cdecl main(void)"
    foo();
    FOO();
}