JavaScript is concatenating instead of adding

Question

I am using these in my code:

alert(mult);  //1.114
alert(profit);  //10
price = (mult * profit) + profit;
alert(price);  //11.14 should be 21.14

I also tried price = ((mult*profit) + profit); but got the same result. Instead of doing (1.114*10) + 10; it is doing 1.114*10=11.14. It is not adding the additional 10 at the end.

The `+` operator concatenates its arguments if you give it a string. That means either `mult` or `profit` is a string. Use `parseInt()` and `parseFloat()` to turn a string into a number. — Sidney, Sep 06 '17 at 18:36
This isn't concatenation. It's doing the multiplication without adding (or concatenating). If it was concatenating, the result would be `"11.1410"`. Execute this in the console: `'1.114' * '10' + '10'` — ps2goat, Sep 06 '17 at 18:38
I'm unable to reproduce. Voting to close the question. @raulxbox please edit your question to include a **Complete, Minimal, and Verifiable example**. — Jared Smith, Sep 06 '17 at 18:42

ps2goat · Answer 1 · 2017-09-06T19:04:04.370

It works fine for me like this. It doesn't seem that concatenation is an issue, because otherwise the value would be '11.1410'.

var mult = 1.114;
var profit = 10;
var price = (mult*profit) + profit;
alert(price);//11.14 should be 21.14

If the profit variable (or both variables) is a string, the result would be "11.1410".

var mult = 1.114;
    var profit = '10';
    var price = (mult*profit) + profit;
    alert(price);//11.1410

The result seems unaffected if just the mult value is a string.

 var mult = '1.114';
        var profit = 10;
        var price = (mult*profit) + profit;
        alert(price);//21.14

score 1 · Answer 2 · answered Sep 06 '17 at 18:37

1

The problem is probably that either mult or profit are of type string and not of type number. This can be fixed by explicitly casting.

You can determine the type of your variables in your browser console (i.e. dev tools) with the typeof keyword like this:

typeof foo

If either of your variables are of type string you can convert them to number with parseFloat:

mult = parseFloat(mult);

answered Sep 06 '17 at 18:37

Jonathan.Brink

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typeof mult "undefined" typeof profit "object" this is what I get in console – raulxbox Sep 06 '17 at 18:40
Try running those commands while at a debugger breakpoint, or just execute those lines of code with a `console` statement – Jonathan.Brink Sep 06 '17 at 18:41
I tried and its working mult = parseFloat(mult); but it is not taking decimal points it shows 21.00 not 21.14 – raulxbox Sep 06 '17 at 18:42
You may need to `parseFloat` both variables – Jonathan.Brink Sep 06 '17 at 18:43

DjaouadNM · Accepted Answer · 2017-09-06T18:44:57.380

1

The variables are probably strings, use parseFloat to convert them to numbers :

mult = parseFloat(mult);
profit = parseFloat(profit);
price = (mult*profit) + profit

edited Sep 06 '17 at 18:44

answered Sep 06 '17 at 18:39

DjaouadNM

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Does not answer the question. – Jared Smith Sep 06 '17 at 18:42
it works but it is not taking decimal values it shows 21.00 not 21.14 – raulxbox Sep 06 '17 at 18:43
@raulxbox use parseFloat() as in the updated answer. – DjaouadNM Sep 06 '17 at 18:44

score 0 · Answer 4 · answered Sep 06 '17 at 18:55

I just ran this is in the chrome js console F12 key.

var mult = 1.114, profit = 10;
console.log('mult=', mult);//1.114
console.log('profit=', profit);//10
price = (mult*profit) + profit;
console.log('price=', price);//11.14 should be 21.14

And got these results

mult= 1.114
profit= 10
price= 21.14

JavaScript is concatenating instead of adding

4 Answers4