Unknown array notation javascript

Question

I've come across the following code, I'm having trouble getting my head around it.

selCOption2[i, 'labelname'] = selOption2Arr[i];

a larger excerpt

var selCOption2 = [];
    var stringContent = '';


    jQuery('#txtTypes').attr("value", selOption1);

    for(var i=0; i<selOption2Arr.length; i++) {
      if(selOption2Arr[i] != 'Plain' || selOption2Arr[i] != 'plain') {
        selCOption2[i, 'labelname'] = selOption2Arr[i];
        selCOption2[i, 'keyname'] = keyname+"_"+selOption2Arr[i].toLowerCase()+"_"+selOption3Arr[0].toLowerCase();
        for(var ifm = 0; ifm < proJsonDetails.images.length; ifm++) {   
          if(proJsonDetails.images[ifm].indexOf(selCOption2[i, 'keyname']) > 0) {
            selCOption2[i, 'image'] = proJsonDetails.images[ifm];
          }
        }
      }
    }

This is valid syntax, but not a sensible thing to do. The expression "i, 'labelname'" is a comma operator expression, and the effect is no different from selCOption2['labelname'] = ... (except for the highly unusual case where there might be a getter function on i, which is not the case here). — apsillers
– apsillers, Commented Nov 21, 2014 at 16:55
…and so what you have is equivalent with selCOption2['labelname'] = selOption2Arr[i];. — The Paramagnetic Croissant
– The Paramagnetic Croissant, Commented Nov 21, 2014 at 16:56

Quentin · Accepted Answer · 2014-11-21 16:56:20Z

4

See a reduced test case:

var a = [ 'x', 'y', 'z' ];
var o = {};
var i = 1;
o[i, 'labelname'] = a[i];
console.log(o);

which gives:

{ labelname: 'y' }

The , operator evaluates as whatever is on the right hand side of it.

There doesn't appear to be any point in having i, in that code.

answered Nov 21, 2014 at 16:56

Quentin

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