I typed this into a template function, just to see if it would work:
if (T==int)
and the intellisense didn\'t complain. Is this valid C++? What
No, you can't use if (T == int) and std::cout<<(int)int;
Just to fit your requirement you should use typeid
operator. Then your expression would look like
if (typeid(T) == typeid(int)) {
...
}
Obvious sample to illustrate that this really works:
#include <typeinfo>
#include <iostream>
template <typename T>
class AClass {
public:
static bool compare() {
return (typeid(T) == typeid(int));
}
};
void main() {
std::cout << AClass<char>::compare() << std::endl;
std::cout << AClass<int>::compare() << std::endl;
}
So in stdout you'll probably get:
0
1
No, this is not valid C++.
IntelliSense is not smart enough to find everything that is wrong with code; it would have to fully compile the code to do that, and compiling C++ is very slow (too slow to use for IntelliSense).
Is this what you're trying to do?
if(typeid(T) == typeid(int))
and this?
cout << typeid(int).name();
Your probably didn't even instantiate your template, that's why it compiled.