Why `var a= function(){}` is similar to `function a(){}` in the following case?

Question

I don't understand a part of the accpeted answer of this OP:

Javascript function scoping and hoisting

Writer says:

"

Also, in this instance,

function a() {}

behaved the same as

var a = function () {};

".

What I know is that function expression is different than function declaration, at least for hoisting. Why are they similar in this case?

Answer 1

The difference is that the first line is defined at run-time, whereas second line is defined at parse-time for a script block.

Look at the full answer on stack, here : var functionName = function() {} vs function functionName() {}

Answer 2

Why are they similar in this case?

Because var is hoisted (but not set), like a function declaration is hoisted, meaning that there is an a in the local scope before a = 10; is evaluated, so the global a never gets modified - the identifier lookup finds the local a first so sets that

---

Related parts of the other question

var a = 1;

function b() {
    a = 10;
    return;

    function a() {}
}
b();
alert(a);

Why is a === 1?

---

That the answer was trying to say was that b is equal to

function b() {
    function a() {}
    a = 10;
    return;
}

Similar to

function b() {
    var a = function () {};
    a = 10;
    return;
}

i.e. there is an identifier a defined in b, so

function b() {
    var a = 10;
    return;
}

And now, obviously, the global a will not be modified by b

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Please note that the position of the var isn't really important, just that it's there, the following will produce the same behaviour

function b() {
    a = 10;
    return;
    var a;
}