Java: Environment

What value will be inside the result variable after executing the code?

public static void main(String[] args) {
  var age = 5;
  var result = generate();
}

static int generate() {
  return age + 3;
}

The correct answer is: the code will fail with an error. Because inside the function there is no variable named age, but the function tries to use it.

The function does not have access to variables declared in other functions.

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Consider another example:

public static void main(String[] args) {
  var age = 5;
  var result = generate();
}

static int generate() {
  var age = 10;
  return age + 3;
}

The age variable in main from within generate is not available, it does not interfere with declaring a variable with the same name in generate - but it will be a different variable. The generate function will be under the name of age to see its variable, and the main - its own.
  
In this case, the result will be the number 13....age = 5 does not affect the function code.

Instructions

This task is not directly related to the lesson, it is just another useful exercise in working with functions.

Write a function getAgeDifference, which takes two years of birth and returns a string with an age difference in the form The age difference is 11.

As usual, the function needs to be public static, not just static, so that we can call it from another class.

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Tips

• Recall that in Java there is a Math.abs function that returns the module of the transferred number: for example, Math.abs (-12) will return 12.

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