Question

First I am not sure what is going on in this bitwise operation.
I get code written and supply to other parties as code snippets.

Now if VAR is unsigned 8bit integer (unsigned char) and r is either 0 or 1 or 2 or 4.
Can following be reversed if the value of r is known and resulting value is there.
VAR |= 1 << r; //that is 200 where VAR was 192 and r was 3

For example initial value of VAR is 192 and value of r is 3 *result is 200*.

Now if I have this 200, and I know the value of r that was 3, can I reverse it back to 192 ?

I hope it is most easy one, but I don’t know these bitwise operations, so forgive me.

Thanks

Answer 1

The answer is no. This is because the | (OR) operator is not a one-to-one function.

In other words, there are multiple values of VAR that can produce the same result.

For example:

r = 3;
var0 = 8;
var1 = 0;

var0 |= 1 << r;  //  produces 8
var1 |= 1 << r;  //  produces 8

If you tried to invert it, you wouldn’t be able to tell whether the original value is 0 or 8.

A similar situation applies to the & AND operator.

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From an information-theory perspective:

The operators | and & incur a loss of information and do not preserve the entropy of the data.

On the other hand, operators such as ^ (XOR), +, and - are one-to-one and thus preserve entropy and are invertible.