C ++模板比int-> unsigned转换更好？

Question

I have two functions like the following

template<typename T>
unsigned int myFunction(T myelement)
{

myelement->func();

return 2;
}

void myFunction(unsigned int myelement)
{
}

and I'm using the following code

myFunction(2);

visual studio 2012 is complaining that "int hasn't ->func()". Why isn't it using the unsigned int version?

Answer 1

You are misreading the error message. The compiler doesn't use the function, it is instantiating it to figure out if it is a candidate. You need to disable the instantiation for non-suitable types:

template<typename T>
typename std::enable_if<!std::is_fundamental<T>::value, unsigned int >::type
myFunction( T myelement )
{
    // ...
}

Live example

Answer 2

Try coding

myFunction(2U);

or

myFunction((unsigned)2);

The literal 2 is of int (not unsigned int ) type. And int is not a class or a type having ->func()

And perhaps specialize the template for unsigned instead of defining a myFunction(unsigned int)