Is it possible to use std::string in a constexpr?

2019-01-10 08:22发布

问题:

Using C++11, Ubuntu 14.04, GCC default toolchain.

This code fails:

constexpr std::string constString = "constString";

error: the type ‘const string {aka const std::basic_string}’ of constexpr variable ‘constString’ is not literal... because... ‘std::basic_string’ has a non-trivial destructor

Is it possible to use std::string in aconstexpr? (apparently not...) If so, how? Is there an alternative way to use a character string in a constexpr?

回答1:

No, and your compiler already gave you a comprehensive explanation.

But you could do this:

constexpr char constString[] = "constString";

At runtime, this can be used to construct a std::string when needed.



回答2:

In C++17, you can use string_view:

constexpr std::string_view sv = "hello, world";

A string_view is a string-like object that acts as an immutable, non-owning reference to any sequence of char objects.



回答3:

Since the problem is the non-trivial destructor so if the destructor is removed from the std::string, it's possible to define a constexpr instance of that type. Like this

struct constexpr_str {
    char const* str;
    std::size_t size;

    // can only construct from a char[] literal
    template <std::size_t N>
    constexpr constexpr_str(char const (&s)[N])
        : str(s)
        , size(N - 1) // not count the trailing nul
    {}
};

int main()
{
    constexpr constexpr_str s("constString");

    // its .size is a constexpr
    std::array<int, s.size> a;
    return 0;
}