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Using std::string in Constant Expressions
It is generally not possible to employ std::string in constant expressions. The reason for this is that std::string has a non-trivial destructor, making its use incompatible with the requirement of constant expressions to be resolved at compile time.
C 20 Solution
However, with C 20, a limited exception has been introduced. If a std::string is destroyed before the end of constant evaluation, it can be used in constant expressions. For example:
constexpr std::size_t n = std::string("hello, world").size();
In this instance, the std::string is created and destroyed within the constant expression, so its use is permissible.
Alternative Solution (C 17 and Later)
A practical alternative to std::string for use in constant expressions is std::string_view. A string_view is an immutable, non-owning reference to a character sequence. It provides similar functionality to std::string, but without a destructor, making it suitable for constant expressions:
constexpr std::string_view sv = "hello, world";
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