Can a template function be called with missing template parameters in C++ ?

Question

This is an interview question, which has been done.

Which line has error ?

  #include<iostream>
  template<class T> void foo(T op1, T op2)
  {
      std::cout << "op1=" << op1 << std::endl;
      std::cout << "op2=" << op2 << std::endl;
  }

  template<class T>
  struct sum
  {
      static void foo(T op1, T op2)
      {
              std::cout << "sum=" << op2 << std::endl ;
      }
  };

  int main()
  {
      foo(1,3);  // line1
      foo(1,3.2);  // line2
      foo<int>(1,3);  // line3
      foo<int>(1, '3') ; // line 4
      sum::foo(1,2) ; // line 5  , 

      return 0;
  }

Line 2 has error because the template parameter is not matching the definition. Line 5 has error because the template parameter is missing.

But, Line 1 is an not an error, I do not know why, does not it also miss template parameter ?

Thanks !

Answer 1

It's called type deducition.

On Line 1, the type of T can be deduced because parameters op1 and op2 are both int, making T an int.

Whereas on Line 2, you are passing both an int and a double while the function accepts both parameters as T, the compiler has no clue whether T should be a double or an int.

Line 3 is fine because you specify int specialization and pass ints in as well (making the specialization redundant but perfectly OK).

Line 4 is OK because you declare T to be an int, then casting the char value of '3' to its numeric int value.

Line 5 is an error because you're accessing a function that gets its type from the templated struct it's in, and type deduction only works for functions.