JavaScript 'hoisting' [duplicate]

问题

I came across JavaScript 'hoisting' and I didn't figure out how this snippet of code really functions:

var a = 1;

function b() {
    a = 10;
    return;

    function a() {}
}

b();
alert(a);

I know that function declaration like ( function a() {} ) is going to be hoisted to the top of the function b scope but it should not override the value of a (because function declarations override variable declarations but not variable initialization) so I expected that the value of the alert would be 10 instead of 1!!

回答1:

The global a is set to 1
b() is called
function a() {} is hoisted and creates a local variable a that masks the global a
The local a is set to 10 (overwriting the function a)
The global a (still 1) is alerted

回答2:

It's because the order of compilation/interpretation in this example is somewhat misleading. The function a () {} line is interpreted before any of the rest of the function is executed, so at the very beginning of the function, a has the value of function a () {}. When you reassign it to 10, you are reassigning the value of a in the local scope of function b(), which is then discarded once you return, leaving the original value of a = 1 in the global scope.

You can verify this by placing alert()s or the like in the appropriate places to see what the value of a is at various points.

回答3:

(1) JavaScript does not have block statement scope; rather, it will be local to the code that the block resides within.

(2) Javascript's declaration of variables in a function scope, meaning that variables declared in a function are available anywhere in that function, even before they are assigned a value.

(3) Within the body of a function, a local variable takes precedence over a global variable with the same name. If you declare a local variable or function parameter with the same name as a global variable, you effectively hide the global variable.

you code is same as: (read comment)

<script>
var a = 1;          //global a = 1
function b() {
    a = 10;         
    var a = 20;     //local a = 20
}
b();
alert(a);           //global a  = 1
</script>

reference:
(1) JavaScript Variable Scope:
(2) A Dangerous Example of Javascript Hoisting
(3) Variable scope

So in your code:

var a = 1;          //global a = 1  
function b() {
    a = 10;         
    return;
    function a() {} //local 
}
b();
alert(a);           //global a = 1

回答4:

function declaration function a(){} is hoisted first, hence in local scope a is created.
If you have two variable with same name (one in global another in local), local variable always get precedence over global variable.
When you set a=10, you are setting the local variable a , not the global one.

Hence, the value of global variable remain same and you get, alerted 1

回答5:

When I read the same article you did JavaScript Scoping and Hoisting, I was confused as well because the author never showed what the two opening example codes are interpreted as in the compiler.

Here is example you provided, and the second on the page:

var a = 1;
function b() {
    function a() {} // declares 'a' as a function, which is always local
    a = 10;
    return;
}
b();
alert(a);

and here is the first example on the page:

var foo = 1;
function bar() {
    var foo; // a new local 'foo' variable
    if (!foo) {
        foo = 10;
    }
    alert(foo);
}
bar();

Hope this helps

来源：https://stackoverflow.com/questions/15311158/javascript-hoisting

标签

javascript

hoisting