I execute function like this:
var a = 123;
function f() {
alert(a);
var a = 9;
}
f();
the result is undefined, why this happened? Why it's not 123?
2 Answers
Your function is actually compiled as:
function f() {
var a;
alert(a);
a = 9;
}
because of variable hoisting: https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Statements/var#var_hoisting
So your function redeclares a as a local variable with the value as undefined, alerts it, and then re-sets its value to 9.
At the time of the alert, its value is undefined because of the hoisting.
4 Comments
hh54188
but if I remove the
var a = 9;, the alert could show correctly 123, why it doesn't redeclares a?
Ian
@hh54188 If you remove
var a = 9;, then there is no redeclaration...you removed it...all your function will be is alert(a);. So it looks up the scope chain and finds a in the global scope as 123, so it's evaluated as 123.hh54188
but the redeclare was happened after the
alert, so this means the variable declare is doesn't matter with exetute order?Ian
@hh54188 Right, but if you read my answer, it explains that the variable declaration was hoisted (read the website link I provided too). Technically, your function is technically executed as the function I included in my answer, because of hoisting. So
a is technically redefined at the beginning of the function (before the alert)If you declare a in a function syntax again that becomes a new variable. If you want to use the previous value 123 then you should not have included the var a = 9 statement, since it creates a new variable.
This may explain in detail :
What is the scope of variables in JavaScript?
ais defined within the scope off, just not at the point where you callalert(). Henceundefined.