Way to differentiate between declared variable vs undeclared

Question

I have a problem. I will only declare a variable if that satisfies my condition.

if(//some condition1)
var some1=1;
else if(//some condition1)
var some2=2;

Now

var some3=some1+some2;

if two conditions are met, it is okay to have sum in some3;

if only one condition is met, I want to have either some1 or some2 in some3

How can I achieve this?

Note: Thanks for answers But i want to declare variables only if that condition is met. Is there any other solution?

why not just declare the variable first and then just set their values when it meets the conditions — artsylar, Aug 01 '13 at 08:57
@Loney you check for the if(var == undefined) condition then do the addition of only defined one — Deepak Ingole, Aug 01 '13 at 08:57
@Grallen I mean to say `var == undefined` .Why to check `typeof`.Lets say var1 is undefined then you check `if (var1 == undefined)` add only `var2` and `var3` — Deepak Ingole, Aug 01 '13 at 09:00
What's the harm in declaring variables when the condition isn't met? — Barmar, Aug 01 '13 at 09:01
Remember that because of the way Javascript hoists variable declarations, a declaration in an `if` statement is hoisted into the function, even if the condition isn't met. — Barmar, Aug 01 '13 at 09:03
@captain: I honestly can't remember which issue I converted to this syntax to avoid. — Grallen, Aug 01 '13 at 09:09
The question title doesn't seem to match with your question. To find out whether a variable is declared you have to try to access it: `try { foo; } catch() { /* foo is not declared */ }`. But you seem to be asking something else in your actual question, so please update the title. — Felix Kling, Aug 01 '13 at 09:20
@Lonely why do want to declare variables only if a condition is met? what advantage do you expect? — basilikum, Aug 01 '13 at 09:57
@all : please see http://stackoverflow.com/questions/17991100/way-to-differentiate-between-declared-variable-vs-undeclared-2 — Lonely, Aug 01 '13 at 10:12
@Deepanshu Ya checked, but it is not the question I wanted. I can't able to clearly state my problem so asked another question. Thanks for help — Lonely, Aug 01 '13 at 11:21

Deepanshu Goyal · Answer 1 · 2015-12-02T12:54:48.377

3

var some3=0;

if(//some condition1){
  var some1=1;
  some3 += some1;
}else if(some condition2){
  var some2=2;
  some3 += some2;
}

EDIT: A better option :

var some3=0;           
if(some condition1){
   some3 += 1;
}else if(some condition2){
   some3 += 2;
}

and moreover both conditions wont meet at same time, so, if you want to you can do like below also

var some3=0;           
if(some condition1){
   some3 += 1;
}
if(some condition2){
   some3 += 2;
}

edited Dec 02 '15 at 12:54

answered Aug 01 '13 at 08:57

Deepanshu Goyal

2,738
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@Deepenshu question says variable should be declared only if condition is true – Deepak Ingole Aug 01 '13 at 08:58
Good logic, but you are re-declaring your variables. Could also improve the formatting a bit – musefan Aug 01 '13 at 09:00
@Deepanshu Please check question.It says **declare variables only if that condition is met** – Deepak Ingole Aug 01 '13 at 09:03
3

no need to declare some1 and some2 here. you can just directly add it to some3. – artsylar Aug 01 '13 at 09:10
I think the OP means `defined` and `undefined` rather than `declared` and `undeclared`. With `var some3;` I declare the variable but it is still undefined. Declaring the variable (even inside the if) will declare it locally for the full function scope. – devnull69 Aug 01 '13 at 09:12
@artsylar you don't know if OP needs the separate values later on – basilikum Aug 01 '13 at 09:14
well now I have no idea what is the requirement... what i thought should be the best , I typed it down !!! because checking the variable for undefined and stuff is just so many lines of code for nothng – Deepanshu Goyal Aug 01 '13 at 09:15
Your code is fine ... the point is that the OP didn't know that a) undefined is not the same as undeclared and b) that declaring and defining a variable later in the scope will result in it being declared earlier and defined later as described by basilikum – devnull69 Aug 01 '13 at 09:17

basilikum · Answer 2 · 2013-08-01T09:12:37.727

First of all, you can't declare your variables only if a certain condition is satisfied. JavaScript has function scope which means, that no matter where in your code you write var some1, it will always be hoisted up, to the top of the function scope:

function A() {
   if (...) {
       var a = 1;
   }
}

will be interpreted as:

function A() {
   var a;
   if (...) {
       a = 1;
   }
}

That being said. I think to solve your problem you just need to initialize your variables with 0:

var some1 = 0,
    some2 = 0,
    some3;

if(//some condition1)
    some1 = 1;
else if(//some condition1)
    some2 = 2;

some3 = some1 + some2;

EDIT

Actually, there IS a way to to declare variables only when a condition is satisfied:

if (...) {
    eval("var x = 1;");
}

But I don't really see any need for a "hack" like that and wouldn't advice that method at all.

score 1 · Answer 3 · answered Aug 01 '13 at 08:58

Initialize both variables to 0 first.

var some1 = 0, some2 = 0;
if (condition1) {
    some1 = 1;
}
if (condition2) {
    some2 = 2;
}
var some3 = some1 + some2;

Since you want to allow both conditions to be met and set both variables, you should not use else if. Just two separate if statements.

score 1 · Answer 4 · answered Aug 01 '13 at 09:00

1

if(//some condition1) {
    var some1 = 1;
} else if(//some condition1) {
    var some2 = 2;
}

var some3 = 0;

if(some1 !== undefined) {
    some3 += some1;
}
if(some2 !== undefined) {
    some3 += some2;
}

answered Aug 01 '13 at 09:00

Virus721

8,061
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This seems the right answer to me. Only one suggestion, use the comparison to `null` or using `typeof some1 === 'undefined'`, since comparing to `undefined` is not totally safe – Niccolò Campolungo Aug 01 '13 at 09:01
Why isn't it totally safe ? – Virus721 Aug 01 '13 at 09:05
Because in older environments it was possible to override the value of `undefined` and change it, for example, to `true`, which would break a lot of things. – Niccolò Campolungo Aug 01 '13 at 09:16
Well that would help people getting rid of their obsolete stuff. – Virus721 Aug 01 '13 at 09:17

score 1 · Answer 5 · answered Aug 01 '13 at 09:00

1

How about an object:

var obj = {};
if(//conditon){
   obj.some1 = 1;
}
else if(//condition){
   obj.some2 = 1;
}

var sum3 = 0;
for(var key in obj){
    sum3 += obj[key];
}

answered Aug 01 '13 at 09:00

Akhil Sekharan

12,467
7
40
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score 0 · Answer 6 · answered Aug 01 '13 at 08:58

0

why not just declare the variable first and then just set their values when it meets the conditions

var some1 = 0;
var some2 = 0;

if (//some condition)
{
  some1 = 1;
}
if (// some condition)
{
  some2 = 2;
}

var some3 = some1 + some2;

answered Aug 01 '13 at 08:58

artsylar

2,648
4
34
51

score 0 · Answer 7 · answered Aug 01 '13 at 08:58

0

Declare all the variables in the begin. Initialize them to 0. If the condition is met, change the variable value. Then, when you make the sum, any value plus 0 remains the same. Or if the condition was met, the value won't be 0.

answered Aug 01 '13 at 08:58

Oscar

13,594
8
47
75

score 0 · Answer 8 · answered Aug 01 '13 at 09:02

0

If you need to use that model, you could just use a fourth variable.

var some0 = 0
if(/*some condition1*/) {
var some1=1;
some0 = some1 + some3
}
else if(/*some condition1*/) {
var some2=2;
some0 = some2 + some3
}
some3=some0

answered Aug 01 '13 at 09:02

Paul

3,634
1
18
23

Naveen Kumar Alone · Answer 9 · 2013-08-01T09:44:33.403

0

var some3 = 0; //Assingning global var.
if(//some condition1)
  var some1=1;
else if(//some condition1)
  var some2=2;

In Conditions

if(some1 != null && some2 != null )
  some3=some1+some2;
else if(some1 != null && some2 == null)
  some3=some1;
else if(some1 == null && some2 != null)
  some3=some2;
else if(some1 == null && some2 == null)
  some3=some3;

Then

alert(some3);

edited Aug 01 '13 at 09:44

answered Aug 01 '13 at 09:04

Naveen Kumar Alone

7,536
5
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score 0 · Answer 10 · answered Aug 01 '13 at 09:15

0

How about:

var some3 =
(2 > 4      ? 10:0) +
(5 < 8      ? 20:0) +
(10 == 11   ? 30:0) +
('a' < 'b'  ? 40:0) ;

I admit this does not look pretty.

answered Aug 01 '13 at 09:15

S.D.

29,290
3
79
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Andreas · Answer 11 · 2013-08-01T09:44:48.583

0

you an use the

if(//some condition1) var some1=1;

else if(//some condition1) var some2=2;

and then try catch statements like in my jsfiddle

try{

 var some3=some1+some2;

}
catch (e){
    try{
      var some3=some1;
      document.getElementById("test").innerHTML= some3;

    }catch (e){
        try{
           var some3=some2 ;
           document.getElementById("test").innerHTML= some3;

        }
        catch (e){
        document.getElementById("test").innerHTML="Something went wrong!!!!or no condition satisfied";
        }
    }
}

edited Aug 01 '13 at 09:44

answered Aug 01 '13 at 09:31

Andreas

970
18
28

(The use of exceptions is inefficient, because of the overhead of throwing exceptions, but I was just giving an extra option form the other answers.) more info about the side effects of try catch in an [old question](http://stackoverflow.com/questions/3217294/javascript-try-catch-performance-vs-error-checking-code). In order to answer that question @jon-j created this [fiddle](http://jsfiddle.net/Mct5N/) which proves this. – Andreas Aug 01 '13 at 12:01

Way to differentiate between declared variable vs undeclared

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