问题
I just can't understand why the the a1 = function ?
and where is my value 1 that was passed to the fn(),
whether it was overrwrited by var a ?
the problem look like caused by the same names( var & function) !
function fn(a) {
console.log("a1 = " + a);
var a = 2;
function a() { }
console.log("a2 = " + a);
}
fn(1);
// a1 = function a() { }
// a2 = 2
function fnx(ax) {
console.log("a1 = " + ax);
var ax = 2;
function b() { }
console.log("a2 = " + ax);
}
fnx(1);
// a1 = 1
// a2 = 2
/* it equal to the final version */
function fn(a) {
var a;
a = function() { }
// function hoisting > variable hoisting
console.log("a1 = " + a);
a = 2;
console.log("a2 = " + a);
}
fn(1);
// a1 = function a() { }
// a2 = 2回答1:
I just can't understand why the the a1 = function ?
Function declarations are:
- Hoisted to the top of the function they appear in
- Declare a local variable (with the same name as the function) in the scope of the function they appear in (this isn't relevant because argument definitions do that too)
- Assign themselves as the value of that variable
and where is my value 1 that was passed to the fn(),
Overwritten by the function declaration
whether it was overrwrited by var a ?
The var is ignored because there is already a local variable named a.
The assignment overwrites the function between the two console.log statements.
Your code is effectively the same as:
function fn(a) {
a = function a() { };
console.log("a1 = " + a);
a = 2;
console.log("a2 = " + a);
}
fn(1);
来源：https://stackoverflow.com/questions/41198746/how-to-understand-javascript-in-deep-with-var-scope-closure