为什么调用我的函数返回未定义？

Question

I'm trying to understand why when I call maxofThree on any array, undefined is returned? I'm fairly new to JS so appreciate your help!

function maxofThree (array) {
    for (var i; i < array.length; i++) {
        var largestNumber = 0;

        if (array[i] > largestNumber) {
            largestNumber = array[i];
        }
        return largestNumber;
    }
}

Answer 1

You have to return after the loop, and you have to start i of at zero.

function maxofThree(array) {
    var largestNumber = array[0];
    for (var i=0; i < array.length; i++) {
        if (array[i] > largestNumber) {
            largestNumber = array[i];
        }
    }
    return largestNumber;
}

As a sidenote, you could use Math.max to get the largest number as well

function maxofThree(array) {
    return Math.max.apply(null, array);
}

Answer 2

function maxofThree (array) {
    for (var i; i < array.length; i++) { //the variable i is not initialized i==undefined
       //At each loop you increment i==undefined undefined++ equals to NaN
       //So i at first loop is undefined, at second loop is NaN
       //undefined < any number except 0, NaN < any number is always false, so the loop is executed only one time
        var largestNumber = 0; //At each loop you initialize/reset the variable largestNumber  to 0

        if (array[i] > largestNumber) {
            largestNumber = array[i]; //i is NaN, array[NaN]==undefined
            //So largestNumber is always undefined
        }
        return largestNumber; //You return largestNumber before next loop occurs, you return undefined
        //The next loop never occurs, because you have done a return
    }
}

Fixed code:

function maxofThree (array) {
    var largestNumber = 0; //Initialize largestNumber to 0 before loops occurs
    // I suppose all numbers are positive, otherwise you need to rethink the solution
    for (var i=0; i < array.length; i++) { //Initialize i to 0
        //At each loop i is incremented until i < array.length

        if (array[i] > largestNumber) {
            largestNumber = array[i];
        }
         //The next loop will occurs
    }
    return largestNumber; //At the end of loop you return the largestNumber
}

Answer 3

You didn't initialize i , so by default it's undefined . When you make the comparison i < array.length , undefined is converted to the number NaN (not a number), and so the expression evaluates to false . That means you never enter the loop and you never hit a return statement before completion of the function, so it returns undefined .

You should initialize i to 0 . After that, there are still 2 more problems.

Assigning 0 to largestNumber at the start of the loop means that it's going to be reset to 0 on every iteration. You should move that statement to before the loop.

Also, having the return in the loop would cause the function to return immediately at the end of the first iteration. You should move that statement to execute after the loop.

While we're at it, you should rename the function. It's not like you're really limiting yourself to 3 elements (not that you ever should). Also, array is too generic. They need to be numbers, right? So call them numbers .

Your function should look like this:

function maximum(numbers) {
    var largestNumber = 0;
    for (var i = 0; i < numbers.length; i++) {
        if (numbers[i] > largestNumber) {
            largestNumber = numbers[i];
        }
    }
    return largestNumber;
}