为什么两个if（'0'== false）和if（'0'）在Javascript中评估为true？ [英] why do both if('0'==false) and if('0') evaluate to true in Javascript?

问题描述

因此，据我所知，Javascript中的 if 语句将其条件的结果转换为布尔值，然后执行它，如下所示

So for what I know is that the if statement in Javascript casts the result of its condition to a Boolean, and then executes it likes the following

if(true) {
    // run this
}

if(false) {
    // do not run this
}

这样可行。但如果我这样做：

And that works. But If I do this:

if('0' == false) {
    // We get here, so '0' is a falsy value
}

然后我会期待这个

if('0') {
    // We don't get here, because '0' is falsy value
}

但我得到了

if('0') {
    // We *DO* get here, even though '0' is falsy value
}

那么发生了什么？显然，如果没有检查它的条件是否是真值或假值，而是进行其他转换？

So what's happening? Apparently, if does not check if its condition a truthy or falsy value, but does some other conversion?

推荐答案

这只是 <$ c $的陷阱之一c> == 规则相当复杂。

This is just one of those "gotchas" with the == rules which are rather complex.

比较x == y，其中x和y是值，产生true或false。这样的比较如下进行：

The comparison x == y, where x and y are values, produces true or false. Such a comparison is performed as follows:

（4）如果Type（x）是Number而Type（y）是String，
返回结果比较x == ToNumber（y）。

(4) If Type(x) is Number and Type(y) is String, return the result of the comparison x == ToNumber(y).

（5）如果Type（x）是String而Type（y）是Number，
返回结果比较ToNumber（x）== y。

(5) If Type(x) is String and Type(y) is Number, return the result of the comparison ToNumber(x) == y.

（6）如果Type（x）是布尔值，则返回比较结果ToNumber（x）== y。

(6) If Type(x) is Boolean, return the result of the comparison ToNumber(x) == y.

（7）如果Type（y）是布尔值，则返回比较结果x == ToNumber（y）。

(7) If Type(y) is Boolean, return the result of the comparison x == ToNumber(y).

在这种情况下，这意味着'0'== false 首先被强制执行'0 '== 0 （按规则＃7）然后在第二次传递时，它被强制转换为 0 == 0 （按规则＃ 5）结果为真。

In this case that means that '0' == false is first coerced do '0' == 0 (by rule #7) and then on the second pass through, it is coerced to 0 == 0 (by rule #5) which results in true.

这个特殊情况有点棘手，因为 false~> 0 而不是'0'〜> true （可能是预期的）。但是，'0'本身就是一个真值，并且可以用上述规则解释行为。要在测试期间进行严格的truthy-falsey相等（不同于严格相等）而不在相等期间进行隐式转换，请考虑：

This particular case is somewhat tricky because of false ~> 0 instead of '0' ~> true (as what might be expected). However, '0' is itself a truth-y value and the behavior can be explained with the above rules. To have strict truthy-falsey equality in the test (which is different than a strict-equality) without implicit conversions during the equality, consider:

!!'0' == !!false

（对于所有值：！falsey - > true 和！truthy - > false 。）

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