Why toString(2) is going to n % 2 and return the divisor remainder?

Question

In this case when n is a number parameter in the function. Why is toString(2) going to n % 2 and return the divisor remainder? I thought that toString just return a number into a string.

+n.toString(2)

function toBinary(n) {

  var toBinary2 = +n.toString(2)

  return toBinary2

}
console.log(toBinary(2))

Please provide an example value of `n`, your expected result, and the actual result. — JLRishe, Jan 12 '18 at 13:02
If you provide an argument to [*Number.prototype.toString*](http://ecma-international.org/ecma-262/8.0/#sec-number.prototype.tostring), it's used as the radix (i.e. base). The default is 10, but you can use any integer from 2 to 36, so what you're getting is n base 2. — RobG, Jan 12 '18 at 13:02
@EdgarChe Ok, and what result were you expecting? 10 isn't the remainder of 2 and 2. — JLRishe, Jan 12 '18 at 13:04

score 0 · Answer 1 · answered Jan 12 '18 at 13:04

0

The parameter to Number#toString is a base.

Calling .toString(2) on a number will return that number in binary.

2 in binary is 10, so that is why (2).toString(2) returns 10.

answered Jan 12 '18 at 13:04

JLRishe

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haa ok, so If I put a parameter into toString(2) is going to return the binary number. Thank you so much! – EdgarChe Jan 12 '18 at 13:06

score 0 · Answer 2 · answered Jan 12 '18 at 13:04

Why is toString(2) going to n % 2 and return the divisor remainder?

toString takes radix as parameter and toString(2) converts the number into binary equivalent string.

+Number(2).toString(2) converts the binary equivalent to the number again, but doesn't do binary to decimal conversion.

So, +Number(2).toString(2) => +"10" => 10

Why toString(2) is going to n % 2 and return the divisor remainder?

2 Answers2