Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Return type deduction in C++14 with Examples

  • Difficulty Level : Expert
  • Last Updated : 28 Jan, 2021

In this article, we will discuss Return Type Deduction in C++14. Using an auto return type in C++14, the compiler will attempt to deduce the return type automatically.

Program 1:

Want to learn from the best curated videos and practice problems, check out the C++ Foundation Course for Basic to Advanced C++ and C++ STL Course for foundation plus STL.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

C++14




// C++14 program to illustrate the
// return type deduction
#include <iostream>
using namespace std;
  
// Function to multiply the two
// numbers a and b
auto multiply(int a, int b)
{
  
    // Return the product
    return a * b;
}
  
// Driver Code
int main()
{
    int a = 4, b = 5;
  
    // Function Call
    cout << multiply(a, b);
  
    return 0;
}
Output:
20

Explanation: In the above program, the multiply(int a, int b) function the compiler will perform the multiplication. As the passed parameters 4 and 5, the compiler will return 20 and since its data type is an integer, the compiler will deduce the type as integer automatically and will return 20 as an integer.



Program 2:

C++14




// C++ 14 program to illustrate the
// return type deduction
#include <iostream>
using namespace std;
  
// Function to increment the
// value of a
auto increase(int& a)
{
    // Increment a
    a++;
  
    // Return the updated value
    return a;
}
  
// Driver Code
int main()
{
    int b = 10;
  
    // Function Call
    int& c = increase(b);
  
    // Print the value b and c
    cout << b << c;
    return 0;
}

Output:

Explanation: As it can be seen that in the above program compiler is showing an error. This is because in the function increase(int &a), the compiler will return 11 and since its type is an integer, the compiler will deduce its type as integer and return it, but in main, we are assigning an integer value to an integer reference variable c, that’s why it was showing error.

Now, the above problem can be fixed in two ways:

using auto&:

C++14




// C++14 program to illustrate return
// type deduction using auto&
#include <iostream>
using namespace std;
  
// Function to increment the value
// of a and return the updated value
auto& increase(int& a)
{
    a++;
    return a;
}
  
// Driver Code
int main()
{
    int b = 10;
  
    // Function Call
    int& c = increase(b);
    cout << b << '\n'
         << c;
  
    return 0;
}
Output:
11
11

using decltype(auto):

C++14




// C++14 program to illustrate return
// type deduction using decltype()
#include <iostream>
using namespace std;
  
// Function that increments the value
// of a and return the updated value
decltype(auto) increase(int& a)
{
    a++;
    return a;
}
  
// Driver Code
int main()
{
    int b = 10;
  
    // Function Call
    int& c = increase(b);
  
    cout << b << '\n'
         << c;
  
    return 0;
}
Output:
11
11



My Personal Notes arrow_drop_up
Recommended Articles
Page :