How to distinguish between a variable that is not declared and a varrable that is declared but not be assigned any value?

Question

Both scenario, the typeof the variable will be "undefined". But undeclared variable will raise a exception.

Is there a easy way to handle this?

Answer 1

You may find the question (and my answer) in How to check if a variable or object is undefined? relevant. In general, I view any access to an "undeclared variable" a programming error.

However, this particular case can *only** be detected with the use of detecting for a ReferenceError exception. But, yuck, yuck, yuck! Remember variable declarations are a static lexical construct, ignoring the quirks with property-variables of the global object.

ReferenceError, and now "strict", exist for a reason and I suspect this is an X-Y problem. I do not even recommend the use of typeof for this purpose: fix the code :-)

Happy coding.

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*It has been pointed out that "variable" in window will also [and only] work for global "variables" (they are really just properties that don't need be qualified in all contexts).

Answer 2

if property we want to check in object whether that it exists or not, even if its undefined.

we will use one of these: 'prop' in obj(to check for properties from prototype chain) or obj.hasOwnProperty('prop')

we need to use methods above to check if property exists as accessing property that has not been declared in object will also return undefined.

var o={};
o.c=undefined;
o.c===undefined; //is true
o.a===undefined; //is true as well even though c exists while a doesn't

commonly not a problem as nobody really declare undefined properties much, but when do so do it like this.

o.c=''; //when it can be string or 
o.c=null; //to clearly indicate that its nothing.
then
o.c === undefined will return false!
note!!!
null ==  undefined //true while
null === undefined //false that's why use three equals to test

For variables not declared and are not inside object. When accessed The compiler will return (reference)error. If it doesn't it means its being treated as a global property, window object property, and was not declared, at least in all parent scope, so it will be undefined just as o.a was at top. it will become window.prop.

so x; //error
but x=3; //no error assumed to be global object.
just like o.abcd = 3; would...
make(declare) a property abcd in object o valued(assigned) 3 all at once.

To avoid properties to become a global variable we use var keyword inside function, like this var k;

One thing you can do about this catch the reference error when throw for a variable that doesn't exist and are thought to be a variable itself.

try {
x
} catch(e){//code to run when x is not declared let alone defined.}

Answer 3

You can try:

var a;
try {
    a;
    alert('a');
} catch(e) { /* a not defined */ }
try {
    b;
    alert('b');
} catch(e) { /* b not defined */ }
alert('done');

DEMO

Answer 4

You should never be attempting to access undeclared vars if you're writing clean JS. To avoid such pitfalls (among many others) start LINTing your JS with http://www.jslint.com/ or http://jshint.com/ .

A good read to help you understand the LINT tools and reasoning behind their findings is Crockford's Book, JavaScript: The Good Parts ( http://www.amazon.com/gp/aw/d/0596517742 ).