change value of several variables in one line javascript

Question

Is there a way to change the value of several variables in one line? The code below is working, but very repetitive. How can I change the values of i, j, k in a more elegant way? Also is it possible to write an if function for all three variables instead of repeating it three times.

var i = 0;
var j = -1;
var k = -2;

function fn() {
   // magic happening;

    i += 1;
    j += 1;
    k += 1;

    if (i > 3) {
        i = 0;
    }
    if (j > 3) {
        j = 0;
    }
    if (k > 3) {
        k = 0;
    }

    // more stuff happening
}

Use a comma? i += 1, j += 1, k += 1; - Also, you could use the ternary operator to make it shorter i = (i > 3) 0 : ++i. — evolutionxbox
– evolutionxbox, Commented Sep 6, 2016 at 8:47
try this var i = 0, j = -1, k = -2; and i += 1, j+=1, k+=1 — Raunak Kathuria
– Raunak Kathuria, Commented Sep 6, 2016 at 8:48

Bhushan · Accepted Answer · 2016-09-06 08:57:04Z

1

Try this:

var i = 0;
var j = -1;
var k = -2;

function fn() {
    i = (i <= 2) ? ++i : 0;
    j = (j <= 2) ? ++j : 0;
    k = (k <= 2) ? ++k : 0;
}

Or by writing another function:

var i = 0;
var j = -1;
var k = -2;

function fn() {
    i = calculateIt(i);
    j = calculateIt(j);
    k = calculateIt(k);
}

function calculateIt(value) {
    return (value <= 2) ? ++value : 0;
}

edited Sep 6, 2016 at 8:57

answered Sep 6, 2016 at 8:50

Bhushan

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1 Comment

softwarenewbie7331 Over a year ago

[(i+1)>3] should become [i > 2]

Nina Scholz · Accepted Answer · 2016-09-06 08:59:46Z

0

You could use a single expression for one assignment.

i = i > 2 ? 0 : i + 1;

You can make an function which keeps the interval.

function incAndLimit(v) {
    return v > 2 ? 0 : v + 1;
}

// usage
i = incAndLimit(i);

edited Sep 6, 2016 at 8:59

answered Sep 6, 2016 at 8:49

Nina Scholz

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Comments

Sandip - Frontend Developer · Accepted Answer · 2016-09-06 09:00:13Z

0

Assign value like this:

i++,j++,k++;

edited Sep 6, 2016 at 9:00

answered Sep 6, 2016 at 8:53

Sandip - Frontend Developer

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1 Comment

Aron Over a year ago

This ignores the event where i,j,k are higher than 3.

JussiR · Accepted Answer · 2016-09-06 09:03:09Z

0

var [i, j, k] = [0, -1, -2]
                .map(v => v + 1)
                .map(v => v > 3 ? 0 : v);

Note that this code requires ES6 compatible environment.

answered Sep 6, 2016 at 9:03

JussiR

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2 Comments

Aron Over a year ago

This only works for declaring the variables, it doesn't work for changing their values once they've been created.

JussiR Over a year ago

You can always redeclare them with var. But yea, there might be a cleaner way depending on what you want to do inside the code. The functional way would be to chain all actions together so you need to make only one declaration.

Aron · Accepted Answer · 2016-09-06 10:13:11Z

0

If you can put the variables in an array you can use ES2015's .map and lambdas:

var list = [0, -1, -2] // replaces separate i, j, and k
list = list.map(x => x++)
           .map(x => x > 3 ? 0 : x);

You can combine these two functions

list = list.map(x => {
    x++;
    return x > 3 ? 0 : x
});

Even if you can't put these variables in an array I'd suggest putting the checking function in a separate function:

function abv3(x) {
    x++;
    return x > 3 ? 0 : x;
}

i = abv3(i), j = abv3(j), k = abv3(k);

edited Sep 6, 2016 at 10:13

answered Sep 6, 2016 at 9:00

Aron

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Comments

Redu · Accepted Answer · 2016-09-06 20:04:07Z

0

I guess this is it by double ES6 array destructuring

var  i = 0,
     j = -1,
     k = -2,
update = ([x,y,z]) => [++x>3?0:x, ++y>3?0:y, ++z>3?0:z];

[i,j,k] = update([i,j,k]);
console.log(i,j,k);
[i,j,k] = update([i,j,k]);
console.log(i,j,k);
[i,j,k] = update([i,j,k]);
console.log(i,j,k);
[i,j,k] = update([i,j,k]);
console.log(i,j,k);
[i,j,k] = update([i,j,k]);
console.log(i,j,k);
[i,j,k] = update([i,j,k]);
console.log(i,j,k);
[i,j,k] = update([i,j,k]);
console.log(i,j,k);
[i,j,k] = update([i,j,k]);
console.log(i,j,k);
[i,j,k] = update([i,j,k]);
console.log(i,j,k);
[i,j,k] = update([i,j,k]);
console.log(i,j,k);
[i,j,k] = update([i,j,k]);
console.log(i,j,k);
[i,j,k] = update([i,j,k]);
console.log(i,j,k);
[i,j,k] = update([i,j,k]);
console.log(i,j,k);

edited Sep 6, 2016 at 20:04

answered Sep 6, 2016 at 19:53

Redu

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