CODEWITHSUNDEEP
c#booleanboolean-operations
trailmax
trailmax

Reputation: 35126

Is there an XNOR (Logical biconditional) operator in C#?

I could not find an XNOR operator to provide this truth table:

a  b    a XNOR b
----------------
T  T       T
T  F       F
F  T       F
F  F       T

Is there a specific operator for this? Or I need to use !(A^B)?

Upvotes: 204

Views: 83212

Answers (5)

Braden Best
Braden Best

Reputation: 8998

XOR A B is the same thing as AND (OR A B) (NAND A B), or in other words, A or B, but not both. XOR A B is only true when A and B are not equal, so it can be also thought of as an inequality check--if XOR A B is true, then A and B must have opposing values (true and false or vice versa).

XNOR is the exact inverse, and can be thought of as an equality check--if XNOR A B is true, then A and B must have the same value (both false or both true).

However, non-boolean cases present problems, like in this example:

a = 5
b = 1

if (a == b) {
   ...
}

This will compare false because 5 = 1 is not true. If your intent is to check that both a and b are the same in terms of being zero/nonzero, then you could do it like this:

if ((a && b) || (!a && !b)) {
    ...
}

or alternately

if (!(a || b) && (a && b)) {
    ...
}

Though this is a really roundabout approach. Remember the intent is to check that both are zero/nonzero. The most direct way to check that would probably be use the Logical NOT operator:

if (!a == !b) {
    ...
}

Upvotes: 5

Peter
Peter

Reputation: 1

I there is a few bitwise operations I don't see as conventional in the whole discussion. Even in c, appending or inserting to the end of a string, all familiar in standard IO and math libs. this is where I believe the problem lies with not and Xnor, not familiar with python but I propose the following example

function BitwiseNor(a as integer)
    l as string
    l=str(a)
    s as string
    For i=len(l) to 0
        s=(str(i)+"="+Bin(i)) //using a string in this example because binary or base 2 numnbers dont exists in language I have used
    next i
endfunction s

function BitwiseXNor(a as integer,b as integer)
    r as integer
    d as integer
    c as string
    c=str(BitwiseOr(a,b))
    r=(val(c,2)) //the number to in this conversion is the base number value
endfunction r

Upvotes: -1

user7797323
user7797323

Reputation: 1

You can use === operator for XNOR. Just you need to convert a and b to bool.

if (!!a === !!b) {...}

Upvotes: -11

Keith Thompson
Keith Thompson

Reputation: 263577

XNOR is simply equality on booleans; use A == B.

This is an easy thing to miss, since equality isn't commonly applied to booleans. And there are languages where it won't necessarily work. For example, in C, any non-zero scalar value is treated as true, so two "true" values can be unequal. But the question was tagged , which has, shall we say, well-behaved booleans.

Note also that this doesn't generalize to bitwise operations, where you want 0x1234 XNOR 0x5678 == 0xFFFFBBB3 (assuming 32 bits). For that, you need to build up from other operations, like ~(A^B). (Note: ~, not !.)

Upvotes: 341

Griffin
Griffin

Reputation: 14654

No, You need to use !(A^B)

Though I suppose you could use operator overloading to make your own XNOR.

Upvotes: 4

Related Questions