关于函数声明的困惑 [英] Confusion about function declaration
问题描述
如果我将一个命名函数分配给一个变量,为什么我无法访问指定的函数:
var a = function b(){
console.log(Hello World);
}
a()// hello world;
b()// b未定义
b // b未定义
在这个时候,我不能删除 a
变量,但是我可以删除 b
,当我删除 b
, a
的函数仍然可以在那里
delete a // false
delete b // true
a()// Hello World
a
,而不是复制,那么为什么 a
函数仍然存在?
如果使用命名函数表达式(不是函数声明!),函数的名称只能在函数内部访问。
从规范(标识符是指函数的名称,即函数标识符(){}
):
可以从 FunctionExpression 的 FunctionBody 中引用ssion ,以允许函数递归调用自身。然而,与 FunctionDeclaration 不同, FunctionExpression 中的标识符不能被引用,也不会影响包含 FunctionExpression 的作用域。
>
但我可以删除
b
不,你不能。当您传递任何不是对 delete
的引用的引用或者引用无法解析时,它会返回 true
。
这是一篇关于所有函数定义内容的精彩文章: http://kangax.github.io / nfe / 。
If I assign a named function to a variable, why I can't access to the named function:
var a = function b() {
console.log("Hello World");
}
a() // hello world;
b() // b is not defined
b // b is not defined
and at this time, I can not delete the a
variable, but I can delete b
, when I delete the b
, the a
's function can still there
delete a //false
delete b //true
a() //Hello World
the b
function isn't just referenced by the a
, not copy, so why the a
function is still there?
If you use a named function expression (not a function declaration!), the name of the function is only accessible form inside the function.
From the specification ("Identifier" refers to the name of function, i.e. function Identifier() {}
):
The Identifier in a FunctionExpression can be referenced from inside the FunctionExpression's FunctionBody to allow the function to call itself recursively. However, unlike in a FunctionDeclaration, the Identifier in a FunctionExpression cannot be referenced from and does not affect the scope enclosing the FunctionExpression.
but I can delete
b
No, you can't. When you pass anything that is not a reference to delete
or the reference cannot be resolved, it will return true
.
This is a great article about all the function definition stuff: http://kangax.github.io/nfe/.
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