C++ Fundamentals for Professionals/

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- Example

- Example

Here, we'll see an example of how decltype can be used to automatically deduce the return type of a function.

We'll cover the following...

The add function #

C++
#include <iostream>
#include <typeinfo>
template<typename T1, typename T2>
auto add(T1 first, T2 second) -> decltype(first + second){
    return first + second;
}
int main(){
  std::cout << std::endl;
  std::cout << "add(1, 1)= " << add(1, 1) << std::endl;
  std::cout << "typeid(add(1, 1)).name()= " << typeid(add(1, 1)).name() << std::endl;
  std::cout << std::endl;
  std::cout << "add(1, 2.1)= " << add(1, 2.1) << std::endl;
  std::cout << "typeid(add(1, 2.1)).name()= " << typeid(add(1, 2.1)).name() << std::endl;
  std::cout << std::endl;
  std::cout << "add(1000LL, 5)= " << add(1000LL, 5) << std::endl;
  std::cout << "typeid(add(1000LL, 5)).name()= " << typeid(add(1000LL, 5)).name() << std::endl;
  std::cout << std::endl;
}

Explanation #

The example implements the add function, which takes two arguments and returns their sum. The return type of the function is deduced by the compiler by applying the decltype operator on the sum of the arguments.

The expression, typeid(add(1, 2.1)).name(), such as in line 19, returns a string representation of the return type.

In the next lesson, we’ll be presented with an exercise related to automatic return type deduction.