在编译时将 const char* 转换为 const char_type*

Question

Consider the following code:

using char_type = /*implementation defined*/;

void foo(const char_type*);

int main()
{
    foo("Hello World!");
}

The string literal "Hello World!" is a const char* that, depending on the implementation, may not be convertible to const char_type* . I want my code to be portable between different implementations so I thought I could define a literal to convert one char after another (this type of conversion is guaranteed to work):

consteval const char_type* operator"" _s(const char*, size_t);

and then use it like this foo("Hello World!"_s) . However, the only implementation I can think of uses new to allocate space and std::copy but that would be extremely slow. I want to make the conversion at compile time and luckily I can use c++20 and the consteval keyword to ensure that the call to the function always produces a contant expression (user defined literals are still normal functions, they can be called at runtime). Any idea how to implement this?

Answer 1

This conversion is possible through a two-step process: first, by declaring a class that can convert a const char * to a char_type array within a compile-time constructor; second, by using that class within a user-defined literal:

#include <algorithm>

template<std::size_t N>
struct string_convert {
    char_type str[N] = {};

    consteval string_convert(const char (&s)[N]) {
        std::copy(s, s + N, str);
    }
};

template<string_convert SC>
consteval auto operator ""_s()
{
    return SC.str;
}

This interface allows for the following use:

void foo(const char_type *s);

foo("Hello, world!"_s);

Try it on Godbolt . Note that neither string_convert nor the user-defined literal appear in the disassembly; all that remains is the converted array.