Explicit Template Argument Specification

[subi09]

i started reading about templates and i got confused on the below.

template<class T>
T max(T t1, T t2)
{
   if (t1 > t2)
      return t1;
   return t2;
}

int main(){

std::cout<<max< int>120,14.55);//=>as per my understanding here if i explicitly mention only one data type,other would be deducted from the argument type(14.55) by the compiler.That means T max(T t1, T t2) instantiates T max(int t1,double t2)in this case.But when i compiled it i got the below warning


warning: passing double for argument 2 to T max(T, T) [with T = int].Then why this warning came is my question since i have already instantiated for T max(int t1,double t2).

also since we are using only T,if i mention max<int>,then T max(T t1, T t2) should be like this int max(int t1, int t2) na??

return 0;

}

Please clear my doubts.else i cant proceed further

[weaknessforcats]

Your template only uses T as a parameter. That means you can't use int and double.

Since the template only returns T you can't return int or double.

A template is only a pattern. When you specialize the template, the compiler makes a copy of it and replaces the T with the type you specify. This creates the function that will be called.

You use explicit specialization then certain types require different logic:

template <class T>
bool IsZero(T arg)
{
     if (arg) return true;
	 return false;
}
//explicit 
//do not use the template when T is double
bool IsZero(double arg)
{
	if (fabs(arg) < 0.0001) return false;
	return true;
}

Ths tells the compiler to not use the template if T is double.

Since bool IsZero(double arg) is an actual function, it will not be in a header file but will be in an implementation file somewhere in the build. Therefore, you need to tip off the compiler that the function exists. You do it this way:

template <class T>
bool IsZero(T arg)
{
     if (arg) return true;
	 return false;
}
//explicit specialization
//do not use the template when T is double
template<> bool IsZero(double arg);
}