Question

A summation of the Explicit Template Argument Specification

template<class T>
T max(T t1, T t2)
{
   if (t1 > t2)
      return t1;
   return t2;
}
max<double>(120, 14.55);   we explicitly determine the type of T as double 

I understand the part above.

Below , it is a bit different

 template<class T>
T SumOfNumbers(int a, int b)
{
   T t = T(); // ???

   t = T(a)+b;  //??? 

   return t;
}

Which takes two ints, and sums them up. Though, summing them in int itself is appropriate, this function template gives opportunity to calculate the sum (using operator+) in any type as required by caller. For example, the get the result in double, you would call it as:

double nSum;
nSum = SumOfNumbers<double>(120,200);    //  ???

I understand the topic “Explicit Template Argument Specification”. But , here the condition is different , bcs function template’s arguments’ types are already is definite.

I can’t understand the lines before the sign “???” ?

Could you please explain it to me step by step ? What does happen at this line

nSum = SumOfNumbers<double>(120,200); 

Does 120 converted 120.0 namely from int to double ?

What T(a) ? What does it mean?

Reference:
http://www.codeproject.com/Articles/257589/An-Idiots-Guide-to-Cplusplus-Templates-Part-1

Answer 1

T t = T();

Initialises t by value-initialisation. For built-in arithmetic types, it is given the value zero; for user-defined types, it is initialised using the default constructor.

(Pedantically, it’s initialised by copying or moving a value-initialised temporary, so this will fail if no copy or move constructor is available; in practice the copy or move will be elided).

t = T(a)+b;

Converts a to type T, adds b to that converted value, and assigns the result to t. If T is a built-in type, then T(a) will use a standard conversion or cast; if it’s user-defined, then it will use a constructor of the form T(int).

There’s no point to the first line, since t is going to be reassigned immediately. The function could be written more clearly as return T(a)+b;

nSum = SumOfNumbers<double>(120,200);

This instantiates the function template with a return type of double, and calls. The overall effect is the same as nSum = double(120) + 200; or nSum = 220.0.