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Java

Posted by Alexander Davison
Alexander Davison
65,469 Points

My version of Hangman (In Java)

After finishing the Java Objects course, I decided to take my Hangman game a little further. I thought it's a little boring if you know the word (since you yourself pass in the word), so I added a flavor of randomness.

However... I was also looking for a Java function that can pick a random value out of an array. I couldn't find a function like that, so I did this:

String[] words = {
    "elephant", "giraffe", "kangaroo", "apples", "orange",
    "laptop", "dictionary", "static", "microsecond",
    "leather", "printer", "station", "picture", "computer",
    "community", "tracks", "trains", "support", "organization",
    "members", "private", "increment", "calculator", "headphones",
    "stickers", "mathematics", "structure", "cryptography",
    "watch", "palace", "castle", "lighter", "scratch", "diamond",
    "emerald"
};

Random rand = new Random();

String word = "";
for (String i : words) {
    if (rand.nextBoolean()) {
        word = i;
    }
}

I know that my method of retrieving a random value won't always work, so I'd appreciate any help on getting a random element out of my array of words.

Thanks! That's all I can say.

Again, any help is deeply appreciated :)

~Alex

1 Answer

codebyamir
codebyamir
12,108 Points

Hey Alexander,

You can use the nextInt() method in the Random class to generate a random array index that will always be within the bounds of the array. The for each loop is not needed here.

Random rand = new Random();
int randomIndex = rand.nextInt(words.length);
System.out.println(words[randomIndex]);
Alexander Davison
Alexander Davison
65,469 Points

Thank you very much!

By the way, I've been experimenting of jshell and I found that rand.nextInt() makes a number too big. I didn't know you can pass in a argument!

Thanks again for letting me accomplish my goal.

~Alex