Variable values are populating incorrectly while multiple declaration

Question

I am learning JavaScript basics. While doing multiple variable declaration in a single statement, getting different results.

var jhon, kate = " kate";
console.log(jhon + kate);
var jhon = "jhon ", kate = " kate";
console.log(jhon + kate);

Question : Why first console.log prints value for kate not for jhon?

You might want to check this question for declaring multiple variable. stackoverflow.com/questions/694102/… — Gelxxxx
– Gelxxxx, Commented Jul 22, 2019 at 5:03
If you want to declare multiple , use = not , var jhon = kate = " kate"; — hs-dev2 MR
– hs-dev2 MR, Commented Jul 22, 2019 at 5:04
@hs-dev2MR Try it under 'use strict'. It only declares jhon, not kate. — melpomene
– melpomene, Commented Jul 22, 2019 at 5:08

Jack Bashford · Accepted Answer · 2019-07-22 05:00:56Z

4

Because doing this:

var jhon, kate = " kate";

Is equivalent to:

var jhon;
var kate = " kate";

Which is:

var jhon = undefined;
var kate = " kate";

Which, when concatenated, gives:

undefined kate

You simply haven't given jhon a value in the first example.

answered Jul 22, 2019 at 5:00

Jack Bashford

44.3k11 gold badges56 silver badges84 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

melpomene · Accepted Answer · 2019-07-22 05:09:08Z

0

var jhon,kate = " kate";

You defined 2 variables, but you only set value for kate. Due to the jhon is not set value, when you console.log it, it will be print undefined

edited Jul 22, 2019 at 5:09

melpomene

86.2k8 gold badges96 silver badges155 bronze badges

answered Jul 22, 2019 at 5:07

Dng

1317 bronze badges

Comments

kcsujeet · Accepted Answer · 2019-07-22 05:13:05Z

-1

what you want is

var jhon = kate = " kate";
console.log(jhon + kate);
var jhon = "jhon ", kate = " kate";
console.log(jhon + kate);

answered Jul 22, 2019 at 5:13

kcsujeet

5512 silver badges12 bronze badges

7 Comments

melpomene Over a year ago

var jhon = kate = "kate"; is wrong because it only declares jhon, not kate.

kcsujeet Over a year ago

got it. but since, what the question is dealing with string and not with arrays or objects, this is also correct right???

melpomene Over a year ago

Do you know what var does?

kcsujeet Over a year ago

detonates a bomb?

melpomene Over a year ago

No, it declares a variable. Please refer to MDN var. In particular, var x = y declares x, but not y, whereas var x, y declares both x and y.

|

Mayank Dudakiya · Accepted Answer · 2019-07-22 05:13:10Z

-1

You should use = to assign a value to the variable not ,

var jhon = kate = "kate";
console.log(jhon + kate);

// Output will be : kate kate

Following approach is easily solve your issue which very easy to handle.

let
  jhon = 'kate',
  kate = 'kate',
  cena = 'foobar'
;

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment

For following a will equal 10 and b will equal 20.

var a, b;
[a, b] = [10, 20];

console.log(a);
// expected output: 10

console.log(b);
// expected output: 20

answered Jul 22, 2019 at 5:13

Mayank Dudakiya

3,88940 silver badges37 bronze badges

1 Comment

melpomene Over a year ago

var jhon = kate = "kate"; is wrong because it only declares jhon, not kate. Destructuring assignment is irrelevant. If you want to assign different values, separate statements are better: a = 10; b = 20;. If you want to assign the same value, use a = b = 42. But you can't combine that with a declaration.

Collectives™ on Stack Overflow

Variable values are populating incorrectly while multiple declaration

4 Answers 4

Comments

Comments

7 Comments

1 Comment

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

Comments

7 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Linked

Related