JS makes me angry - if statement, & operator

Discussion in 'JavaScript' started by Icanthackthis, Jul 16, 2018.

0
  1. #1
    Hi guys!

    I have this simple code:

    
//$UT = parseInt(document.getElementById("XML_type").value);
//$ASZ = document.getElementById("XML_person").value;
$UT = 1;
$ASZ = "0123456789101";

$LE= $ASZ.length;
                             
if(($UT&2 == 2 && $LE != 13) || ($UT&2 != 2 && $LE> 0)){
   msg = "ALERT";
   console.log(msg);
}
    Code (markup):
    I just simply cant understand what's going on...
    If $LE= 13 there is no way to go to the statement. I think i strongly misunderstand something :(
    Help plz!
     
    Icanthackthis, Jul 16, 2018 IP
  2. sarahk

    sarahk iTamer Staff

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    #2
    What is the $UT&2 meant to be checking? Looks like string concatenation but then comparing to an integer

    I had a mess about changing the & to a +

    var UT = 1;
var ASZ = "0123456789101";
//var ASZ = "012345678910";
var LE = ASZ.length;

console.log(UT & 2);
console.log(LE);

if ((UT + 2 == 2 && LE != 13) || (UT + 2 != 2 && LE > 0)) {
  msg = "ALERT";
  console.log(msg);
}

if (UT + 2 == 2 && LE != 13) {
  msg = "ALERT1";
  console.log(msg);
}
else if(UT + 2 != 2 && LE > 0) {
  msg = "ALERT2";
  console.log(msg);
}
    Code (JavaScript):
     
    sarahk, Jul 16, 2018 IP
  3. Icanthackthis

    Icanthackthis Peon

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    #3
    Thanks for your reply!
    I got it!

    The correct way to put the & operator into '()', as both are logical expressions, and == has the priority over the &. Grrr

    
//$UT = parseInt(document.getElementById("XML_type").value);
//$ASZ = document.getElementById("XML_person").value;
$UT = 1;
$ASZ = "0123456789101";

$LE= $ASZ.length;

if((($UT&2) == 2 && $LE != 13) || (($UT&2) != 2 && $LE> 0)){
msg = "ALERT";
console.log(msg);
}
    Code (markup):
     
    Icanthackthis, Jul 16, 2018 IP
  4. deathshadow

    deathshadow Acclaimed Member

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    #4
    Since you're only evaluating binary and, you don't need to == 2 in the first bloody place. You're only checking one binary bit, so it's either going to be 0 or 2. With zero being boolean false, you could simply have written that as:

    
if (
	(($UT & 2) && $LE != 13) ||
	(!($UT & 2) && $LE > 0)
) {
    Code (markup):
    Since we have loose boolean, use it. Same as doing a "test" in assembler, jz or jnz.

    Also if LE would never be negative, you could ditch the > 0 for just && $LE

    Though this isn't PHP, what's with all the dollar sign prefixing?
     
    Last edited: Jul 19, 2018
    deathshadow, Jul 19, 2018 IP
  5. sarahk

    sarahk iTamer Staff

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    #5
    I feel better now for not knowing either!
     
    sarahk, Jul 19, 2018 IP