I want the following code to output "test" instead of "X" for the case when using std::string
using the same initialization as the other basic types. std::string
now calls the constructor with an initializer_list
and therefore the template specialization of get
for char
is called.
#include <sstream>
#include <string>
#include <iostream>
// Imagine this part as some kind of cool parser.
// I've thrown out everything for a simpler demonstration.
template<typename T> T get() {}
template<> int get(){ return 5; }
template<> double get(){ return .5; }
template<> char get(){ return 'X'; }
template<> std::string get(){ return "test"; }
struct Config {
struct proxy {
// use cool parser to actually read values
template<typename T> operator T(){ return get<T>(); }
};
proxy operator[](const std::string &what){ return proxy{}; }
};
int main()
{
auto conf = Config{};
auto nbr = int{ conf["int"] };
auto dbl = double{ conf["dbl"] };
auto str = std::string{ conf["str"] };
std::cout << nbr << std::endl; // 5
std::cout << dbl << std::endl; // 0.5
std::cout << str << std::endl; // 'X'
}
Is there a nice way of doing this without breaking the consistent look of the variable initializations?
std::string
has a constructor that takes an initializer_list<char>
argument; that constructor will always be considered first when you use list-initialization with a non-empty braced-init-list, that's why the char
specialization of get()
is being matched.
If you use parentheses instead of braces for all the initializations, the initializer_list
constructor will no longer be the only one considered in the std::string
case.
auto nbr = int( conf["int"] );
auto dbl = double( conf["dbl"] );
auto str = std::string( conf["str"] );
However, this change alone doesn't work because you have an implicit user-defined conversion template that can yield any type. The code above, in the std::string
case, results in matches for all std::string
constructors that can be called with a single argument. To fix this make the conversion operator explicit
.
struct proxy {
// use cool parser to actually read values
template<typename T>
explicit operator T(){ return get<T>(); }
};
Now, only the explicit conversion to std::string
is viable, and the code works the way you want it to.