Solution Review: Max with Nested Functions

Let’s go over the solution for the challenge: Max with Nested Functions.

We'll cover the following

Task

In this challenge, you had to create a nested function max which would help its parent function mainMax to compute the maximum of three numbers.

Solution

A skeleton of the mainMax function was already provided for you. Let’s look it over.

int mainMax(int a, int b, int c) {
  
}

mainMax takes three parameters of type int and returns a value of type int.

Let’s go over the step-by-step process for writing the max function.

  • max is intended to break down the bigger problem into a smaller one. While mainMax returns the maximum of three numbers, max returns the maximum of two of them. This means that it will take two parameters of type int and return the greater of the two. To find the maximum of two numbers, a simple if-else expression can be used.
int max(int x, int y) {
    if(x > y){
      return x;
    } else{
      return y;
    }  
 }
  • As for the return value of mainMax, we simply needed to call the max function. The first argument will be one of the three numbers passed to mainMax and the second argument will be the maximum of the remaining two. To get the second argument, we will use the max function again as it returns the maximum of two numbers.
return max(a,max(b,c));

You can find the complete solution below:

You were required to write the code given from line 1 to line 10.

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