JavaScript - 防止八进制转换

问题描述

我将数字输入作为参数，只是试图解释前导零.但似乎 javascript 在我可以对数字做任何事情之前将数字转换为八进制.到目前为止，解决它的唯一方法是，如果我最初将数字作为字符串传递，但我希望在传递后有另一种方法来转换它?到目前为止尝试过(使用 017 提醒我注意八进制行为):

I'm taking a numerical input as an argument and was just trying to account for leading zeroes. But it seems javascript converts the number into octal before I can do anything to the number. The only way to work around it so far is if I pass the number as a string initially but I was hoping there'd be another way to convert it after it is passed? So far tried (using 017 which alerted me to the octal behaviour):

017.toString(10) // 15
parseInt(017,10) // 15
017 + "" //15
new Number(017) //15

new Number('017') //17
parseInt('017', 10) // 17

给定

function(numb) {

if (typeof numb === number) {
// remove leading zeroes and convert to decimal
}
else {
 // use parseInt
}

}

'use strict' 似乎也没有像一些较早的帖子所建议的那样解决这个问题.有什么想法吗?

'use strict' also doesn't seem to solve this as some older posts have suggested. Any ideas?

推荐答案

1. 如果你采用数字输入"，你应该始终保证有一个字符串.在这种情况下，我知道没有返回 Number 的输入法.由于您收到一个字符串，因此 parseInt(.., 10) 将始终足够.017 只有在源代码中按字面意思写成八进制时才被解释为八进制(或者当缺少 parseInt 的 radix 参数时).

2. 如果出于某种奇怪的原因，您最终将十进制解释为八进制，并且您想将该值反向转换回十进制，这很简单:用八进制表示该值并将其重新解释为十进制:

1. If you take "numerical input", you should always definitely guaranteed have a string. There's no input method in this context that I know that returns a Number. Since you receive a string, parseInt(.., 10) will always be sufficient. 017 is only interpreted as octal if written literally as such in source code (or when missing the radix parameter to parseInt).

2. If for whatever bizarre reason you do end up with a decimal interpreted as octal and you want to reverse-convert the value back to a decimal, it's pretty simple: express the value in octal and re-interpret that as decimal:

var oct = 017; // 15
parseInt(oct.toString(8), 10) // 17

虽然您可能不知道输入最初是否被解释为八进制，但这不是您应该做的事情.

Though because you probably won't know whether the input was or wasn't interpreted as octal originally, this isn't something you should have to do ever.

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