How to concatenate operators in javascript?

Question

I am writing next_previous() function for pagination purpose, I have for loop which is moving from 0 to the given length. I want to use the same loop for two cases from 0 to 10 and from 10 to 0.

for (var i = 0; i < 10; i++) {
}

for (var i = 10; i > 0; i--) {
}

to use both cases in one loop I am doing something like this but not working

var  i = 0; a = '', b = '';

if(somevar === true){
   i = 0 , a = '++', var b = '<';
}else{
   i = 10 , a = '--', var b = '>';
}

for (i; i +b+ 0; i+a) {
}

now problem is javascript not allowing concatenation this way, how can I achieve this?

See screenshot

What are you trying to achieve? Can you post a sample of the output you want? — Eddie
– Eddie, Commented Nov 22, 2017 at 7:30
I don't know any language that allows "concatenation" like this. It looks like you expect + to behave like eval. — Bergi
– Bergi, Commented Nov 22, 2017 at 7:38

Suren Srapyan · Accepted Answer · 2017-11-22 07:36:52Z

3

Try this approach which uses for the logic part ( increment and condition ) functions.

ES6

let i = 0;
let a;
let b;
let count = 0;

let somevar = true;

if(somevar) {
   i = 0;
   count = 10;
   a = () => i++;
   b = () => i < count;
} else {
   i = 10;
   count = 0;
   a = () => i--;
   b = () => i > count;
}

for (; b(); a()) {
   console.log(i);
}

ES5

var i = 0;
var a;
var b;
var count = 0;

var somevar = true;

if(somevar) {
   i = 0;
   count = 10;
   a = function() { i++; };
   b = function() { return i < count; };
} else {
   i = 10;
   count = 0;
   a = function() { i--; };
   b = function() { return i > count; };
}

for (; b(); a()) {
   console.log(i);
}

edited Nov 22, 2017 at 7:36

answered Nov 22, 2017 at 7:31

Suren Srapyan

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6 Comments

Nick Taylor Over a year ago

Suren can you please do this in javascript.

Suren Srapyan Over a year ago

This is Javascript :)

Nick Taylor Over a year ago

I think it's ECMA SCRIPT or something

Suren Srapyan Over a year ago

This is just ES6 specification

Nick Taylor Over a year ago

one little question Suren?

|

Bergi · Accepted Answer · 2017-11-22 07:43:47Z

3

It seems like you are looking for an eval solution, but this really is not how one would approach this problem. Rather go for a functional design:

function forward(cb) {
    for (var i = 0; i < 10; i++) cb(i);
}
function backward(cb) {
    for (var i = 10; i > 0; i--) cb(i);
}

const loop = somevar ? forward : backward;
loop(i => {
    …
});

answered Nov 22, 2017 at 7:43

Bergi

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Comments

doubleOrt · Accepted Answer · 2017-11-30 19:10:28Z

This answer is a less-lazy version of @BiswajitPanday's answer. I would have edited his answer instead, but I realized that it's been a week since this question was posted, so an edit on his answer wouldn't be as fruitful since OP wouldn't be notified

Here is an eval solution:

function loopFoo(backwardsDirection, callback) {
  if(backwardsDirection === true) {
  var i = 10, b = '>', c = 0, a = '--';
  } else {
  var i = 0, b = '<', c = 10, a = '++';	
  }

  for (i; eval("i" + b + c); eval("i" + a)) {
  callback(i);	
  }
}


console.log("Forwards:");
loopFoo(false, console.log);

console.log("Backwards:");
loopFoo(true, console.log);

console.log("Forwards:");
loopFoo(false, console.log);

That function (eval) is generally frowned upon though, I would go with @Bergi's solution, if not that, @Suren's, and if not his either, my solution.

Biswajit Panday · Accepted Answer · 2017-11-22 07:44:14Z

-1

Just use eval(i+b+10) instead of i+b+10

answered Nov 22, 2017 at 7:44

Biswajit Panday

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3 Comments

Bergi Over a year ago

But i is 0 not "i".

Biswajit Panday Over a year ago

Some int + some string produces a string. As b is string so i will be also a string after concatenate.

Bergi Over a year ago

Yes, but it produces the wrong string to be evaluated, at least in the case eval(i+a).

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