在JavaScript中，为什么“0”等于假，但当通过“if”测试时，它本身不是假的?

如下所示，Javascript中的"0"为false:

>>> "0" == false
true

>>> false == "0"
true

那么下面为什么打印“哈”呢?

>>> if ("0") console.log("ha")
ha

当前回答

这是按规格的。

12.5 The if Statement 
.....

2. If ToBoolean(GetValue(exprRef)) is true, then 
a. Return the result of evaluating the first Statement. 
3. Else, 
....

根据规范，ToBoolean是

抽象操作ToBoolean根据表11将其参数转换为Boolean类型的值:

这个表格是这样描述字符串的:

如果参数为空String(其长度为零)，则结果为假; 否则结果为真

现在，为了解释为什么"0" == false，您应该读取相等运算符，它表示它从抽象操作GetValue(lref)中获得其值，与右侧的相同。

将相关部分描述为:

if IsPropertyReference(V), then 
a. If HasPrimitiveBase(V) is false, then let get be the [[Get]] internal method of base, otherwise let get
be the special [[Get]] internal method defined below. 
b. Return the result of calling the get internal method using base as its this value, and passing 
GetReferencedName(V) for the argument

或者换句话说，字符串有一个基元，它会回调内部的get方法，结果看起来是false。

如果你想用GetValue运算来求值，请使用==，如果你想用ToBoolean运算，请使用===(也称为“严格”相等运算符)

2011-09-30 19:47:30

其他回答

原因是，当显式执行"0" == false时，两边都被转换为数字，然后执行比较。

如果执行:if ("0") console.log("ha")，则正在测试字符串值。任何非空字符串为真，而空字符串为假。

Equal (==) If the two operands are not of the same type, JavaScript converts the operands then applies strict comparison. If either operand is a number or a boolean, the operands are converted to numbers if possible; else if either operand is a string, the other operand is converted to a string if possible. If both operands are objects, then JavaScript compares internal references which are equal when operands refer to the same object in memory. (From Comparison Operators in Mozilla Developer Network)

2011-09-30 19:46:02

显示问题的表格:

和= =

这个故事的寓意使用===

表生成credit: https://github.com/dorey/JavaScript-Equality-Table

2011-09-30 19:36:43

这都是因为ECMA规格…"0" == false，因为这里指定的规则http://ecma262-5.com/ELS5_HTML.htm#Section_11.9.3…if('0')的结果为true，因为这里指定的规则http://ecma262-5.com/ELS5_HTML.htm#Section_12.5

2011-09-30 20:10:04

The "if" expression tests for truthiness, while the double-equal tests for type-independent equivalency. A string is always truthy, as others here have pointed out. If the double-equal were testing both of its operands for truthiness and then comparing the results, then you'd get the outcome you were intuitively assuming, i.e. ("0" == true) === true. As Doug Crockford says in his excellent JavaScript: the Good Parts, "the rules by which [== coerces the types of its operands] are complicated and unmemorable.... The lack of transitivity is alarming." It suffices to say that one of the operands is type-coerced to match the other, and that "0" ends up being interpreted as a numeric zero, which is in turn equivalent to false when coerced to boolean (or false is equivalent to zero when coerced to a number).

2011-09-30 19:49:41