Returning values from a function in C++

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Functions return a value or return void. Void is a signal to the compiler that no value will be returned.

To return a value from a function, write the keyword return followed by the value you want to return. The value might itself be an expression that returns a value. For example:

return 5;
return (x > 5);
return (MyFunction());

These are all legal return statements, assuming that the function MyFunction() itself returns a value. The value in the second statement, return (x > 5), will be zero if x is not greater than 5, or it will be 1. What is returned is the value of the expression, 0 (false) or 1 (true), not the value of x.

When the return keyword is encountered, the expression following return is returned as the value of the function. Program execution returns immediately to the calling function, and any statements following the return are not executed.

It is legal to have more than one return statement in a single function. Program below illustrates this idea.

A demonstration of multiple return statements.

1:     // Demonstrates multiple return
2:     // statements
3:
4:     #include <iostream.h>
5:
6:     int Doubler(int AmountToDouble);
7:
8:     int main()
9:     {
10:
11:         int result = 0;
12:         int input;
13:
14:         cout << "Enter a number between 0 and 10,000 to double: ";
15:         cin >> input;
16:
17:         cout << "\nBefore doubler is called... ";
18:         cout << "\ninput: " << input << " doubled: " << result << "\n";
19:
20:         result = Doubler(input);
21:
22:         cout << "\nBack from Doubler...\n";
23:         cout << "\ninput: " << input << "   doubled: " << result << "\n";
24:
25:
26:         return 0;
27:    }
28:
29:    int Doubler(int original)
30:    {
31:         if (original <= 10000)
32:              return original * 2;
33:         else
34:              return -1;
35:         cout << "You can't get here!\n";
36: }
Output: Enter a number between 0 and 10,000 to double: 9000

Before doubler is called...
input: 9000 doubled: 0

Back from doubler...

input: 9000   doubled: 18000

Enter a number between 0 and 10,000 to double: 11000

Before doubler is called...
input: 11000 doubled: 0

Back from doubler...
input: 11000  doubled: -1

Analysis: A number is requested on lines 14 and 15, and printed on line 18, along with the local variable result. The function Doubler() is called on line 20, and the input value is passed as a parameter. The result will be assigned to the local variable result, and the values will be reprinted on lines 22 and 23.
On line 31, in the function Doubler(), the parameter is tested to see whether it is greater than 10,000. If it is not, the function returns twice the original number. If it is greater than 10,000, the function returns -1 as an error value.

The statement on line 35 is never reached, because whether or not the value is greater than 10,000, the function returns before it gets to line 35, on either line 32 or line 34. A good compiler will warn that this statement cannot be executed, and a good programmer will take it out!



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