javascript - 为什么parseInt（1 / 0,19）返回18？(Why does parseInt(1/0, 19) return 18?)

Question

I have an annoying problem in JavaScript.

(我在JavaScript中遇到了烦人的问题。)

> parseInt(1 / 0, 19)
> 18

Why does the parseInt function return 18 ?

(为什么parseInt函数返回18 ？)

  ask by cebor translate from so

Answer 1

The result of 1/0 is Infinity .

(1/0的结果是Infinity 。)

parseInt treats its first argument as a string which means first of all Infinity.toString() is called, producing the string "Infinity" .

(parseInt将其第一个参数视为字符串，这意味着首先调用Infinity.toString() ，生成字符串"Infinity" 。)

So it works the same as if you asked it to convert "Infinity" in base 19 to decimal.

(因此它的工作原理与您要求它将基数19中的"Infinity"转换为十进制相同。)

Here are the digits in base 19 along with their decimal values:

(以下是基数19中的数字及其小数值：)

Base 19   Base 10 (decimal)
---------------------------
   0            0
   1            1
   2            2
   3            3
   4            4
   5            5
   6            6
   7            7
   8            8
   9            9
   a            10
   b            11
   c            12
   d            13
   e            14
   f            15
   g            16
   h            17
   i            18

What happens next is that parseInt scans the input "Infinity" to find which part of it can be parsed and stops after accepting the first I (because n is not a valid digit in base 19).

(接下来发生的是parseInt扫描输入"Infinity"以找到它的哪个部分可以被解析并在接受第一个I之后停止（因为n不是基数19中的有效数字）。)

Therefore it behaves as if you called parseInt("I", 19) , which converts to decimal 18 by the table above.

(因此，它的行为就像你调用了parseInt("I", 19) ，它被上面的表转换为十进制18。)