当提供未声明的变量标识符时，typeof如何绕过ReferenceError？
[英]How does typeof circumvent the ReferenceError when supplied an undeclared variable identifier?
问题描述
foo; // ReferenceError: foo is not defined
typeof(foo); // undefined
How does typeof circumvent the ReferenceError when supplied an undeclared variable identifier? 
当提供未声明的变量标识符时， typeof如何绕过ReferenceError ？ Is this just JavaScript interpreter "magic" or can it be explained in terms of user-land concepts? 这只是JavaScript解释器“神奇”还是可以用用户土地概念来解释？
1 个解决方案
解决方案1
7 已采纳 2015-03-19 22:53:37
No, this cannot be explained in user-land concepts - it's "magic" if you want. 不，这在用户土地概念中无法解释 - 如果你愿意，这是“神奇的”。
EcmaScript uses the Reference specification type to explain cases like this. EcmaScript使用Reference规范类型来解释这样的情况。 These references are used to describe the semantics of assignments, method calls , eval and many more. 这些引用用于描述赋值， 方法调用 ， eval等语义。 Typically, the GetValue algorithm is called on them to dereference them (eg in the evaluation of your expression statement), and this does throw the ReferenceError when the reference is not resolvable. 通常，在它们上调用GetValue算法以取消引用它们（例如，在表达式语句的求值中），并且当引用不可解析时，这会抛出ReferenceError 。
The typeof operator in contrast does not just do GetValue, but has a special case to handle these undeclared-variable references: 相比之下， typeof运算符不只是执行GetValue，而且还有一个特殊情况来处理这些未声明的变量引用：
If
Type(val)is Reference , thenType(val)是Reference ，那么a.
IsUnresolvableReference(val)istrue, return"undefined".IsUnresolvableReference(val)为true，则返回"undefined"。
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