What does this construct do {flag = flag | a;}?

I’m trying to reuse some old code I wrote and forget where I found the syntax(I’m no expert - probably fit better in the ‘dabbler’ category).

if(pmIn != ""){flag = flag | a;}

Is this a short cut for the ternary operator? As in

flag = (pmIn != "") ? flag : a


if(pmIn != ""){flag = flag;} else {flag = a;}


No. It would be if it had || between the two values though instead of the single |.

One | by itself is a bitwqise or operator and so the value returned from flag | a will be a value with bits set on where ever either of those two values had the bit set on.

Thanks, now I remember what I was doing. Here’s the whole snippet

//get combinations of times
  var flag = 0;
  var a = 1;
  var b = 2;
  var c = 4;
	var d = 8;

  if(amIn != ""){flag = flag | a;}
  if(amOut != ""){flag = flag | b;}
  if(pmIn != ""){flag = flag | c;}
  if(pmOut != ""){flag = flag | d;}

    case 3:
      bb = cTFD_03(amIn, amOut);	//, hrs);
    case 9:
      cTFD_09(amIn, pmOut);	//, hrs);
    case 12:
      cTFD_12(pmIn, pmOut);	//, hrs);
    case 15:
      cTFD_15(amIn, amOut, pmIn, pmOut);	//, hrs);

So when different combinations of the if’s are true it produces different values for flag, then I use that value in the switch statement.
Just proves what a waste of time comments are. :slight_smile:

The added complexity doesn’t provide much of any benefit either.
Have a look at the following, which performs the same job in an easier to understand manner:

if (amIn && amOut && pmIn && pmOut) {
    cTFD_15(amIn, amOut, pmIn, pmOut);
} else if (amIn && amOut) {
    cTFD_03(amIn, amOut);
} else if (amIn && pmOut) {
    cTFD_09(amIn, pmOut);
} else if (pmIn && pmOut) {
    cTFD_12(pmIn, pmOut);

Then it’s just a matter of giving the functions suitably descriptive names.

Also bitwise operators in JavaScript are extremely inefficient - unlike most other languages where they are amongst the fastest operations, in JavaScript they are the slowest by a long way.

Thanks for the speedy help and tips everyone. I’ll be sure to try here again next time I forget what I’ve done (or need help doing something I’ve never done.)