Question

Write the definition of a void function that takes as input an integer and outputs two times the number if it is even; otherwise it outputs five times the number.

Answer 1

Answer:

Written in C++

void number(int n){

if(n%2 == 0)

cout<<2 * n;

else

cout<<5 * n;

}

Explanation:

The programming language is not stated.

However, I answered using C++

This line defines the function as void

void number(int n){

This line checks if the number is even

if(n%2 == 0)

If yes, it doubles the number and prints the output

cout<<2 * n;

If otherwise,

else

It multiplies the number by 5 and prints the output

cout<<5 * n;

}

To call the function from main, use:

number(n);

Note than n must be declared as integer

See attachment

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