Why the first code works and the second does not work?

Question

I want to know how JavaScript deals with code, what happens in the browser
Code 1 (working code)

let array = ['Item 1', 'Item 2', 'Item 3'];
array.forEach(function(item) {
    if (item === 'Item 2') {
        item = item.toUpperCase();
    } else {
        item = item.toLowerCase();
    }
    console.log(item);
});

// output item 1
//        ITEM 2
//        item 3

Code 2 (not working)

let array = ['Item 1', 'Item 2', 'Item 3'];
array.forEach(function(item) {
    if (item === 'Item 2') {
        item.toUpperCase();
    } else {
        item.toLowerCase();
    }
    console.log(item);
});

// output Item 1
//        Item 2
//        Item 3

second code ise not assign item. it must be like item = item.toUpperCase(); — hsyn.ozkara, Nov 13 '18 at 10:26
@NinaScholz that's just part of it. The point is that `toUpperCase` doesn't modify the string in place, but that would be the case even if strings were mutable. — Federico klez Culloca, Nov 13 '18 at 10:26
Please perform a [minimum investigation](https://www.google.nl/search?q=javascript+toUpperCase+does+not+change+site:stackoverflow.com) — mplungjan, Nov 13 '18 at 10:28

score 2 · Accepted Answer · answered Nov 13 '18 at 10:25

2

item.toUpperCase(); returns the uppercased string that is why when you do item = item.toUpperCase(); the uppercased string is assigned to item and since item is a array element, the element is changed by its reference.

answered Nov 13 '18 at 10:25

Ankit Agarwal

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score 0 · Answer 2 · answered Nov 13 '18 at 10:26

0

Here item = item.toUpperCase(); in bold is creating a local variable which is getting printed , but in the later case it is just the reference and the function argument which is printing

answered Nov 13 '18 at 10:26

brk

48,835
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In both case `item` is local variable but first saved changes and second doesn't – Mohammad Nov 13 '18 at 10:29

Why the first code works and the second does not work?

2 Answers2