I don't understand why only num1 would print output when they both should be. Am I missing something here?
var num1 = 0x200127;
var num2 = 0x200124;
if(num1 & 0x100 == 0x100){
console.log("num1: " + (num1 & 0x100 ) );
}
if(num2 & 0x100 == 0x100){
console.log("num2: " + (num2 & 0x100 ) );
}This is an issue with order of operations. For reference, check the table of JavaScript operator precedence.
== has a precidence of 10 while & has a precidence of 9, so == gets evaluated first.
So your code is essentially evaluating:
num & (0x100 == 0x100)
Which is equivalent to:
num & true
num1 is outputted while num2 is not because:
0x200127 & true == 1 (true)
0x200124 & true == 0 (false)
Try putting your bitwise operations in parenthesis, as the grouping operator has the highest precedence of all.
if((num1 & 0x100) == 0x100){
console.log("num1: " + (num1 & 0x100 ) );
}
if((num2 & 0x100) == 0x100){
console.log("num2: " + (num2 & 0x100 ) );
}
Test it below:
var num1 = 0x200127,
num2 = 0x200124,
output = document.getElementById('output');
if ((num1 & 0x100) == 0x100) {
output.innerHTML += "<p>num1: " + (num1 & 0x100) + "</p>";
}
if ((num2 & 0x100) == 0x100) {
output.innerHTML += "<p>num2: " + (num2 & 0x100) + "</p>";
}<div id="output"></div>The == operator has a higher precedence than &. So
x & y == z
is evaluated as
x & (y == z)
In the second case, that makes the condition evaluated to 0 and thus false:
num2 & 0x100 == 0x100
0x200124 & 0x100 == 0x100
0x200124 & true
0x200124 & 1
0
You want to change the precedence by using the grouping operator:
if((num1 & 0x100) == 0x100){
console.log("num1: " + (num1 & 0x100 ) );
}
if((num2 & 0x100) == 0x100){
console.log("num2: " + (num2 & 0x100 ) );
}
I found the answer, I guess that == has a higher precedence than the & operator. if I change the code to this, it works.
var num1 = 0x200127;
var num2 = 0x200124;
if(num1 & 0x100 == 0x100){
console.log("num1: " + (num1 & 0x100 ) );
}
if((num2 & 0x100) == 0x100){
console.log("num2: " + (num2 & 0x100 ) );
}In javascript == has higer precedence order then bitwise AND operator thus on comparison of 0x100 == 0x100 always result in 1 that is true, and
if we do Bitwise AND of 1 with even number, result will always be 0.
And if we do Bitwise AND of 1 with odd number, result will always be 1
.
