Reputation:
#include <iostream>
template<typename T = char>
T cast(T in){
return in;
}
int main(){
std::cout << cast<>(5) << std::endl;
return 0;
}
Above will print 5
instead of an empty character, as the function is supposed to return a character by default and not an int. What am I ding wrong?
Edit:
Forcing it with std::cout << cast<char>(5) << std::endl;
shows an empty character.
Upvotes: 0
Views: 40
Reputation: 9668
The declaration of 5
is an integer by default. This causes 'T' to be overridden with the type int, rather than using your default type. If you really wanted the char of value 5 (which you probably don't), you could specify it as '\x5'
.
For the ascii character 5....
int main(){
std::cout << cast('5') << std::endl;
return 0;
}
Default types in templates tend to be useful when it's not easy to determine the template type, e.g. cast from int
template<typename T = char>
T cast(int v){
return T(v);
}
and now this will default to a method that casts an int to a char (rather than a int to int).
std::cout << cast(53) << std::endl;
Upvotes: 1