C++ String Template Argument. I was playing with strings in templates. Is it possible in modern c++ (c++17 or greater) to pass a string literal as a parameter to a c++ template?
String Concatenation in C++
Template < > // [2] struct named_type < integer > {using. Web modified 8 months ago. Web as mentioned in other answers, a string literal cannot be used as a template argument. Web template < fixed_string > // [1] struct named_type {}; Is it possible in modern c++ (c++17 or greater) to pass a string literal as a parameter to a c++ template? Web a constraint is a sequence of logical operations and operands that specifies requirements on template. Web to match a template template argument a to a template template parameter p, p must be at least as. Web the usage of std::decay_t will cause the type of abc (which is char const (&) [4]) to decay to char const *. Web stringify template arguments. Web #include template struct type_string_t { static constexpr const char data[sizeof.(chars)] = {chars.};
Web as far as i know, you cannot pass a string literal in a template argument straightforwardly in the current standard. Web the usage of std::decay_t will cause the type of abc (which is char const (&) [4]) to decay to char const *. Web modified 8 months ago. Web as far as i know, you cannot pass a string literal in a template argument straightforwardly in the current standard. Web to match a template template argument a to a template template parameter p, p must be at least as. Web template < fixed_string > // [1] struct named_type {}; Is it possible in modern c++ (c++17 or greater) to pass a string literal as a parameter to a c++ template? Web as mentioned in other answers, a string literal cannot be used as a template argument. Web variadic templates can also be used to create functions that take variable number of arguments. Web #include template struct type_string_t { static constexpr const char data[sizeof.(chars)] = {chars.}; Web with c++17, you can use std::from_chars, which is a lighter weight faster alternative to std::stof and std::stod.it doesn't involve.
