How do i do a clean boolean add in javascript?
1+1 = 0;
1+0 = 1;
1+1+1 = 1;
etc. can one just sum booleans?
true+true = false
false+true = true;
etc.
4 Answers
Just use bitwise XOR operator:
1 ^ 1 = 0
1 ^ 0 = 1
1 ^ 1 ^ 1 = 1
FWIW: The same works for most high-level programming languages.
5 Comments
Aadit M Shah
This is not the same as boolean addition. Boolean addition is the xor operator.
basilikum
@AaditMShah I think the equivalent to a boolean addition really is the OR operator. Look at this for example: allaboutcircuits.com/vol_4/chpt_7/2.html
Aadit M Shah
@basilikum I understand what you're trying to say but I believe that the OP is looking to create an adder in JavaScript. Hence the sum is found using xor and the carry is found using and. See: en.wikipedia.org/wiki/Adder_(electronics)
basilikum
@VisioN yes that's right. But OR still seems to be equivalent to a boolean addition. Just google "boolean addition". However, if OP really wants
1+1 to be equal to 0, than he probably doesn't really look for a boolean addition, since this would give 1 as a result.
David Karlsson
I actually ended up doing just (bool1+bool2+bool3)%2
What you're looking for is the xor operator:
1 ^ 1 = 0;
1 ^ 0 = 1;
1 ^ 1 ^ 1 = 1;
Comments
1 ^ 1 = 0;
1 ^ 0 = 1;
for boolean this can be achieved by using Short-circuit AND and OR operators.
function myXOR(a,b) {
return ( a || b ) && !( a && b );
}
myXOR(true,true) == false
Comments
you can code yor for bool like this:
function xor(a,b) {
if (a === true && b === true) {
return false;
}
return a || b;
}
OR in typescript:
xor(a: boolean,b: boolean): boolean {
if (a === true && b === true) {
return false;
}
return a || b;
}
1+1supposed to be 1 in boolean addition?0 + 0 = 0,0 + 1 = 1,1 + 0 = 1and1 + 1 = 0 (with a carry over of 1). Hence boolean addition is the xor operator and the carry can be found using boolean multiplication (i.e. the and operator).1+1equals1.andand boolean addition is in factor. However I believe that the OP is looking to create a half adder. This requires the xor operation.