How to join variable-length parameters as a string - c++

Question

I want to pass variable-length parameters in the function, concatenate these parameters with commas, and finally return the form of the string with parentheses. If the parameter type is a string or char *, then automatically With single quotes, how does this work? Thank you!
For example: join (1, 2, "hello", 3, "world") returns the string "(1, 2, \"hello\", 3, \"world\")"

Answer 1

You could create some function templates to do the job.
#include <iostream>
#include <sstream>
#include <type_traits>
// a function that takes a variable amount of arguments
template<typename T, class... Args >
std::string join_helper(const T& t, Args...args) {
std::ostringstream ss;
using type = std::remove_cv_t<std::remove_reference_t<T>>;
// check if " should be added
if constexpr(std::is_same_v<type, const char*> ||
std::is_same_v<type, std::string> ||
std::is_same_v<type, std::string_view>)
{
ss << '"';
}
ss << t; // stream out the current value
if constexpr(std::is_same_v<type, const char*> ||
std::is_same_v<type, std::string> ||
std::is_same_v<type, std::string_view>)
{
ss << '"';
}
// do we have more arguments? if so, add ", " and call join_helper again
if constexpr (sizeof...(args) > 0) {
ss << ", ";
ss << join_helper(args...);
}
return ss.str();
}
// the function you will use that adds ( and ) around the return value from join_helper
template<class... Args>
std::string join(Args...args) {
if constexpr(sizeof...(args) > 0)
return '(' + join_helper(args...) + ')';
else
return "()";
}
int main() {
std::cout
<< join(1, 2.3, "hello", 4, std::string("world")) << '\n'
<< join() << '\n'
;
}
Output:
(1, 2.3, "hello", 4, "world")
()