CODEWITHSUNDEEP
javascriptif-statement
LiamOakley
LiamOakley

Reputation: 11

Why does this if statement work but this other solution does not?

Im doing this for a dice game where the player is switched if either dice rolled is a 1 or if 6 is rolled twice in a row. My friends code worked but mine didn't, It looks like my if statement accomplishes the same thing.

This code works(friends code):

if (dice === 1 || diceTwo === 1) {
    nextPlayer();
} else if (doubleSix === 6) {
    if (dice === 6 || diceTwo === 6) {
        roundScore = 0;
        score[activePlayer] = 0;
        nextPlayer();
    }
} else {
    roundScore += (dice + diceTwo);
    document.querySelector('#current-' + activePlayer).textContent = roundScore;
    doubleSix = dice;
}

This code does not (my code):

if (dice !== 1 || diceTwo !== 1) {
    //Add score
    if (doubleSix === 6 && dice === 6) {
        roundScore = 0;
        score = 0;
        nextPlayer();
    } else if {
        roundScore += (dice + diceTwo) ;
        document.querySelector('#current-' + activePlayer).textContent = roundScore;
        doubleSix = dice;
    }
    } else {
    //Next Player
    nextPlayer();
}

Upvotes: 1

Views: 59

Answers (4)

vogomatix
vogomatix

Reputation: 5041

Firstly your friends code only requires one dice to be a 1. You require both dice to be a 1 to run nextPlayer

This is because of what is called De Morgan's laws

Your code should have

if (dice !== 1 && diceTwo !== 1) {

Suggested Improvements..

As a general rule it is a bad idea to call items of a similar nature dice, diceTwo etc. It is much better to have an array of dice, as if you increase the number of dice, the code still works without modification.

Also, I am not sure why you are only looking for a six with the first dice of the previous round, joined with any dice of the current round. I would have thought you were looking for any six in the previous round with any six in the current round...

Your friends code would be better as...

var foundSix = false;

// method to sum array
function sum(total, num) {
   return total + num;
}

// ... more code

// check for a 1...
if (dice.indexOf(1) >= 0) {
    nextPlayer();
} else if (foundSix) {
    // check for a 6
    if (dice.indexOf(6) >= 0) {
        roundScore = 0;
        score[activePlayer] = 0;
        nextPlayer();
    }
} else {
    // roundScore += (dice[0] + dice[1]);
    // use array reduce here...
    roundScore = dice.reduce( sum, roundScore);
    document.querySelector('#current-' + activePlayer).textContent = roundScore;
    // doubleSix = dice[0];
    // check ALL dice for a six..
    foundSix = (dice.indexOf(6) >= 0);
}

Upvotes: 3

M3NTA7
M3NTA7

Reputation: 1347

Just a quick look, but shouldn't this: (dice !== 1 || diceTwo !== 1) be changed to an AND like this: (dice !== 1 && diceTwo !== 1) because you are checking that a one was not rolled by both dice and diceTwo?

Upvotes: 0

Daniel Beck
Daniel Beck

Reputation: 21475

Your if statements do not do the same thing.

Your friend's code:

if (dice === 1 || diceTwo === 1) {
    // dice is 1, or diceTwo is 1.
} else { 
    // neither dice nor diceTwo is 1.
}

Your code:

if (dice !== 1 || diceTwo !== 1) {
    // dice is not 1, or dice2 is not 1.  In other words this will match all cases except snake eyes.
} else {
    // both dice and diceTwo === 1
}

Upvotes: 0

Thomas
Thomas

Reputation: 181715

Read up on De Morgan's rule. The negation of dice === 1 || diceTwo === 1 is not dice !== 1 || diceTwo !== 1, but rather dice !== 1 && diceTwo !== 1.

In words: the opposite of "one of the dice is 1" is not "one of the dice is not 1", but rather "both of the dice are not 1".

Upvotes: 1

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