4.6. The return keyword

The return keyword immediately returns from a function. If the function returns a value, then an expression or value may follow the return keyword.

The first example is area, which takes a double as a parameter, and returns the area of a circle with the given radius:

double area (double radius) {
  const double pi = 4*atan (1.0);
  double area = pi * radius * radius;
  return area;
}

The first thing you should notice is that the beginning of the function definition is different. Instead of void, which indicates a void function, we see double, which indicates that the return value from this function will have type double.

Void functions do not need a return statement, but all functions that declare a return value must have at least one return statement, or the program will crash when the function is called.

Note

Functions can have any return type in C++. The return type is always specified before the name of the function.

Also, notice that the last line is an alternate form of the return statement that includes a return value. This statement means, “return immediately from this function and use the following expression as a return value.” The expression you provide can be arbitrarily complicated, so we could have written this function more concisely:

double area (double radius) {
  return 4*atan(1.0) * radius * radius;
}

On the other hand, temporary variables like area often make debugging easier. In either case, the type of the expression in the return statement must match the return type of the function. In other words, when you declare that the return type is double, you are making a promise that this function will eventually produce a double. If you try to return with no expression, or an expression with the wrong type, the compiler will take you to task.

Q-1: What should the return type of the below function be?

________ weather (int temp) {
 string outside = "";
 if (temp < 50) {
   outside = "cold";
 }
 else {
   outside = "warm"
 }
 return outside;
}

• double
• There are no doubles used in this function.
• int
• The parameter does not have to be the same type as the return type.
• string
• The variable "outside" is being returned, which is of string type.
• char
• The return type is actually a string of chars.

Q-2: What will print?

#include <iostream>

int timesTwo(int x) {
  int y = x;
  return y;
  y = y * 2;
}

int main () {
  int i = 2;
  std::cout << timesTwo(i);
  return 0;
}

• 4
• The function returns y before reaching the line where y is doubled.
• 2
• Because the return statement in the timesTwo function returns prior to the modification of y, 2 is returned and then printed.
• 16
• The function returns y before reaching the line where y is doubled.
• The function does not return.
• The function has an integer return type, so it WILL return an int.

Q-3: A variable that exists only inside a function, is called a variable.

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More to Explore

• The return keyword from cppreference.com

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