如何创建带有前置长度字节的编译时常量字符串？

Question

I have some old code that uses #define to, um, define some string literals, eg

#define FredString "\004FRED"

That I would like to update to avoid using #define . Closest I get is something like the following:

static constexpr char* FSraw= "FRED";
static constexpr char FSlen= (char)(sizeof(FSraw) - 1);
static constexpr char* FredString= FSlen FSraw;

but the compiler seems unhappy on the third line.

What would be the best way to construct such a string at compile time? Obviously I could still explicitly encode the length, but also obviously, that's more error prone.

Looking for solutions for C++17 or earlier.

Answer 1

template <std::size_t N>
constexpr std::array<char, N+1> AddLenPrefix(const char (&str)[N])
{
    std::array<char, N+1> ret{};
    ret[0] = N-1; // Exclude the '\0'.
    for (std::size_t i = 0; i < N; i++)
        ret[i+1] = str[i];
    return ret;
}

static constexpr auto FredString = AddLenPrefix("FRED");

Answer 2

You can also have something like:

#include <string_view>
#include <utility>

template <std::string_view const& S, typename>
struct prepend_length_impl;

template <std::string_view const& S, std::size_t ... Is>
struct prepend_length_impl<S, std::index_sequence<Is...>> {
    static constexpr const char value[]{ (sizeof...(Is) + '0'), S[Is]..., 0 };
};

template <std::string_view const& S>
struct prepend_length {
    static constexpr std::string_view value =
        prepend_length_impl<S, std::make_index_sequence<S.size()>>::value;
};

and then use it like:

static constexpr std::string_view raw = "fred";
static constexpr std::string_view fred = prepend_length<raw>::value; // "4fred"

DEMO