Lexical scope in javascript function

Question

The following snippet:

a = 0;

function f1() {
    a = 1;
    f2();
}

function f2() {
    return a;
}

f1();

returns undefined.

From what I understand, functions get access to variables when they're defined, and access the values of those variables when they're being executed. So, in this case I would've guessed f2 to have access to the global variable 'a', and read it's modified value (1). So why is it undefined?

The `return` in `f2` does not return from `f1`… But yes, all three `a`s refer to the same global variable. — Bergi, May 26 '13 at 21:55

score 5 · Accepted Answer · answered May 26 '13 at 21:24

5

You're not returning the result of the invocation of f2(), or anything else, in the f1 function, so f1 is correctly returning undefined.

answered May 26 '13 at 21:24

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good catch! The function gets the value returned by its body, but it still has to return that to its caller. – tldr May 27 '13 at 00:17

score 0 · Answer 2 · answered Oct 27 '14 at 14:43

Perhaps what you were after was the following:

a = 0; // variable a defined in the global scope and set to 0

function f1() {
    a = 1; // since a is declared without var,
           // sets the value of global variable a to 1
    return f2();
}

function f2() {
    return a; // since a was not declared with var,
              // return a from the global scope
}

alert(f1()); // displays the computed value 1 of a

Regards.

Lexical scope in javascript function

2 Answers2