var a = 1;
function b() {
function a() {}; // local scope
a = 10; // global scope
}
b();
alert(a);
It alerts 1 and not 10. I am wondering why is it so?
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1stackoverflow.com/questions/15057649/…Robert– Robert2015-02-02 01:29:32 +00:00Commented Feb 2, 2015 at 1:29
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3 Answers
A function name and a variable are essentially the same thing in Javascript. You can declare a function via:
var a = function () {};
Which is the same as function a() {} for most intents and purposes. Both create a symbol in the current scope and make that symbol's value a function.
What you're doing is you're shadowing the global a with your own local a. It makes no difference whether this local a was defined via var a or function a.
Comments
Your code is the same as this:
var a = 1;
function b() {
var a = function () {}; // local scope
a = 10;
}
b();
alert(a);
Thus, the declaration of your local scope function creates a new local variable named a that initially has a function assigned to it, but you then reassign it to the value of 10. The higher scoped a is not touched by this inner assignment.
If the outer a definition is in the global scope, then you could assign to it with:
window.a = 10;
If it is not in the global scope, then it has been "hidden" by the inner definition of a and there is no way to reach the outer a directly from the inner scope.
Comments
JavaScript is different from other languages:
JavaScript® (often shortened to JS) is a lightweight, interpreted, object-oriented language with first-class functions
What is first-class?
allows functions to be passed around just like any other value.
So as jfriend00 pointed out it converts the function into a variable locally in the function thus not changing the global variable.
