空数组是虚假的，但是[]吗? 0:1等于0 [英] Empty array is falsy, yet [] ? 0 : 1 evaluates to 0

问题描述

如果空数组[]在JavaScript中为 falsy ，则为什么当在三元运算符中用作谓词时，该运算符等于第一个选项?

If the empty array [] is falsy in JavaScript then why, when used as the predicate in the ternary operator, the operator evals to the first option?

console.log([] == false); // prints true
console.log([] ? 0 : 1);  // prints 0 !

推荐答案

首先，一个空数组不是falsey，要使用!!获取它的实际布尔值前缀，例如:console.log(!![]) // true

First, an empty array is not falsey, to get the actual boolean value prefix it with !! such as: console.log(!![]) // true

您从第一次比较中得到的结果是强制性的.通过==操作数进行左右比较时，JavaScript尝试将输入强制为通用类型，如下所示:

The result you got from the first comparison is due to coercion. When doing a left vs. right comparison via the == operand JavaScript attempts to coerce the inputs to a common type, as follows:

1. 发现右侧为Boolean，因此将其转换为Number值，导致比较为[] == 0.
2. 由于左侧是Object，因此将其转换为基本体.在这种情况下，通过[].toString()通过String进行比较，导致比较结果为"" == 0.
3. 由于左侧是String，因此将其转换为number值，从而导致0 == 0.
4. 由于双方都是Number原语，因此将它们的值进行比较，从而得出真实情况

1. The right-hand side is found to be Boolean so it is converted to a Number value, resulting in the comparison being [] == 0.
2. Since the left-hand side is an Object it is converted to a primitive; in this case a String via [].toString(), resulting in the comparison being "" == 0.
3. Since the left-hand side is a String, it gets converted to a number value resulting in 0 == 0.
4. Since both sides are Number primitives, their values are compared, resulting in a truthy condition

使用第二个表达式时，没有右手可以比较，因此它仅测试输入是否为真.数组不是null，undefined，0，""或NaN，因此是正确的.

With the second expression, there is no right-hand to compare to, so it simply tests if the input is truthy. The array is not null, undefined, 0, "" or NaN, therefore, it is truthy.

要在两个输入之间强制进行非强制比较，应使用===操作数.

To force a non-coercive comparison between two inputs you should use the === operand.

Ecma标准第6版在强制性上的条目
JS强制解释了

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