CODEWITHSUNDEEP
javascript
Apurva Bagaria
Apurva Bagaria

Reputation: 27

Why does parseInt return a different result to toFixed?

console.log(typeof(parseInt((0.1 + 0.2).toFixed(1)))); // number
console.log((0.1 + 0.2).toFixed(1) == 0.3); // true
console.log((parseInt((0.1 + 0.2).toFixed(1))) === 0.3); // false

Can someone please explain why the last statement does not return true?

Upvotes: 1

Views: 375

Answers (4)

guyett
guyett

Reputation: 11

Remember that parseInt() converts to an integer, so JavaScript rounds.

So:

  1. parseInt(0.1+0.2) always returns 0. 0 is a number so this returns true.
  2. (0.1 + 0.3) returns 0.3, meaning that this statement is true.
  3. parseInt(0.1 + 0.3) rounds to 0 so this returns false.

Upvotes: 1

CHRIS LEE
CHRIS LEE

Reputation: 786

parseInt will retrieve only integer value. In your case, parseInt(0.3) will return 0 which is not equal to 0.3 Go with Number(0.3) instead of using parseInt.

Upvotes: 0

CertainPerformance
CertainPerformance

Reputation: 370989

parseInt will try to turn its argument into an int (integer). Decimal values will turn into integers. So

(0.1+0.2).toFixed(1)

turns into 0.3, and

parseInt((0.1+0.2).toFixed(1))

turns into 0 (because parseInt floors non-integers, and 0.3 floored is 0).

If you just want to cast to number, use Number instead:

console.log((Number((0.1+0.2).toFixed(1)))===0.3);

Keep in mind that due to floating-point weirdness, 0.1 + 0.2 results in 0.30000000000000004 - that's why calling toFixed on it first, to trim off some of the decimal points, is needed to compare it against 0.3 properly.

Upvotes: 7

Ainaraza
Ainaraza

Reputation: 62

It is because you are using parseInt(0.1 + 0.2). parseInt converts it to an integer and acts like floor, so parseInt(0.3) is giving 0

Upvotes: 1

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