write a simple decode function in javascript

Question

I am a beginner in javascript and I try to write a simple decoding function. Here it is:

function decode(text) {
let letter, count, i, result, letters;
result = "";
for (i = 0; i < text.length; i++) {
        count = text[i];
        if (typeof Number(count) === "number") {               
            count = Number(count);
            console.log(count);  // this line only for testing
            letter = text[i + 1];
            letters = letter.repeat(count);
            result += letters;
            console.log(result);  // this line only for testing
        } else {
            result += text[i];
        }
    }
    return result;
}
console.log(decode("v2rt3b2x"));

I need help in understanding why the if clause is not being skipped when typeof Number(count) === "number" is false. It returns this:

NaN
VM708:12 
VM708:8 2
VM708:12 rr
VM708:8 NaN
VM708:12 rr
VM708:8 NaN
VM708:12 rr
VM708:8 3
VM708:12 rrbbb
VM708:8 NaN
VM708:12 rrbbb
VM708:8 2
VM708:12 rrbbbxx
VM708:8 NaN

As I said, I am a beginner, so please be mercifull...

I got the right answer, but I will try to figure out later why count - 1 works, for now I thank everybody for their help.

function decode(text) {
let letter, count, i, result, letters;
result = "";
for (i = 0; i < text.length; i++) {
    count = parseInt(text[i]);
    if (!isNaN(count)) {               
        //console.log(count);  // this line only for testing
        letter = text[i + 1];
        letters = letter.repeat(count-1);
        //console.log(letters);
        result += letters;
        //console.log(result);  // this line only for testing
    } else {
        result += text[i];
    }
}
return result;
}
console.log(decode("v2rt3b2xz3f2d 2s2 j"));
vrrtbbbxxzfffdd ss  j

typeof Number(n) always return number. Try using if (typeof count === "number") insted of if (typeof Number(count) === "number"). — Sudhir Ojha, Sep 30 '18 at 09:08
If you want to check whether a character (from your string) is a digit, check [this question](https://stackoverflow.com/questions/8935632/check-if-character-is-number) out. — Jeto, Sep 30 '18 at 09:12

aviya.developer · Answer 1 · 2018-09-30T09:11:32.570

3

typeof Number(x) will always be 'number' no matter what x is. You are literally casting whatever x is to a 'number' type.

So the if clause is always true and never gets skipped.

edited Sep 30 '18 at 09:11

answered Sep 30 '18 at 09:09

aviya.developer

3,343
2
15
41

Ahmet Can Güven · Answer 2 · 2018-09-30T09:26:18.153

3

The problem is located here.

typeof Number(count) === "number"

When you try to cast something to number, if it is not possible you will get NaN as the value. And typeof NaN is number.

Read more about NaN

Adding Number.isNaN check to your if statement will be the solution.

function decode(text) {
let letter, count, i, result, letters;
result = "";
for (i = 0; i < text.length; i++) {
        count = text[i];
        var parsedNumber = Number(count);
        if (!Number.isNaN(parsedNumber)) {               
            letter = text[i + 1];
            letters = letter.repeat(parsedNumber);
            result += letters;
            console.log(result);  // this line only for testing
        } else {
            result += text[i];
        }
    }
    return result;
}
console.log(decode("v2rt3b2x"));

edited Sep 30 '18 at 09:26

answered Sep 30 '18 at 09:13

Ahmet Can Güven

5,392
4
38
59

1

This is not how [`Number.isNaN`](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/isNaN) works. It will always return false here. It returns only true when the actual value is `NaN` (the special value of type Number). You want [`isNaN`](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/isNaN) (the "default" one) instead, ideally after using `parseInt` on it, otherwise it will also consider the space character a number. – Jeto Sep 30 '18 at 09:18
Thank you, I am working on this and I see that I get closer to getting the right result – TitianaD Sep 30 '18 at 09:23
@Jeto Well I don't think so. If you look at the implementation of `isNaN`, It just returns if the value can be converted to a number. But` Number.isNaN` returns whether the given value is `NaN`. So, I don't think that we will need `isNaN` for the value we already tried to convert number using `Number(count)` – Ahmet Can Güven Sep 30 '18 at 09:30
@AhmetCanGüven You have edited your code, using `Number.isNaN` on `parsedNumber` is better, but it will still fail for spaces as `Number(' ')` will return `0`, which is a digit. [This thread](https://stackoverflow.com/questions/8935632/check-if-character-is-number) pretty much answers this question. – Jeto Sep 30 '18 at 09:34
@Jeto Ok I see now, I forgot to use `parsedNumber` and used `count`. You were right in that case. – Ahmet Can Güven Sep 30 '18 at 09:36

score 2 · Answer 3 · answered Sep 30 '18 at 09:26

Beside the wrong comparison, you could use an approapriate splitting of the string and map the repeated characters.

By taking

'v2rt3b23x'

and splitting

[
    "v2",
    "r",
    "t3",
    "b23",
    "x"
]

and mapping

[
    "vv",
    "r",
    "ttt",
    "bbbbbbbbbbbbbbbbbbbbbbb",
    "x"
]

you get with join

'vvrtttbbbbbbbbbbbbbbbbbbbbbbbx'

console.log(
    'v2rt3b23x'
        .split(/(?=\D)/)
        .map(v => v[0].repeat(v.slice(1) || 1))
        .join('')
);

write a simple decode function in javascript

3 Answers3