Month: September 2016

Comparing the expressive power of Java with Ruby’s

imrehorvathLeave a comment

I came across this programming exercise on GitHub.

Test 1: FizzBuzzBasic: Write some code that prints out the following for a contiguous range of numbers:

the number
‘fizz’ for numbers that are multiples of 3
‘buzz’ for numbers that are multiples of 5
‘fizzbuzz’ for numbers that are multiples of 15
e.g. if I run the program over a range from 1-20 I should get the following output:

#!console

1 2 fizz 4 buzz fizz 7 8 fizz buzz 11 fizz 13 14 fizzbuzz 16 17 fizz 19 buzz

The Java solution I’ve found at the same location is listed below.

package codingtests.fizzbuzz;

import com.google.common.base.Preconditions;

import java.io.PrintStream;
import java.util.*;
import java.util.function.IntFunction;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class FizzBuzzBasic {

    public static void main(String... arguments) {
        FizzBuzzBasic fb = new FizzBuzzBasic();
        fb.checkCommandLineArguments(arguments);
        fb.convertToFizzBuzzAndOut(Integer.valueOf(arguments[0]), System.out);
    }

    public void convertToFizzBuzzAndOut(int upperBound, PrintStream out){
        getFizzBuzzValuesForNumbers(1,upperBound).stream().forEach(out::println);
    }

    public void checkCommandLineArguments(String... arguments){
        Preconditions.checkArgument(arguments.length == 1, "Program must get exactly one argument");
        try{
            Integer.valueOf(arguments[0]);
        } catch (NumberFormatException ex){
            throw new IllegalArgumentException("Argument must be a valid number");
        }
    }

    public List<String> getFizzBuzzValuesForNumbers(int firstNumber, int lastNumber){
        return IntStream.range(firstNumber,lastNumber+1).mapToObj(this::convertNumberToFizzBuzz).collect(Collectors.toList());
    }

    public String convertNumberToFizzBuzz(int number){
        Preconditions.checkArgument(number > 0,"Number should be greater than 0!");

        if (number % 15 == 0) return "fizzbuzz";
        if (number % 3 == 0) return "fizz";
        if (number % 5 == 0) return "buzz";

        return String.valueOf(number);
    }

}

The Ruby solution I’ve came up with is cited below.

(1..20).lazy.map {|n| if (n % 15) == 0; 'fizzbuzz' elsif (n % 5) == 0; 'buzz' elsif (n % 3) == 0; 'fizz' else n end}.each {|x| print "#{x} "}

Expressing the solution in Ruby is easy and straightforward. It’s actually a one-liner. Easy to read and understand. The Java solution is way more verbose and obscure.

Comparing the expressive power of Java with Ruby’s