CODEWITHSUNDEEP
cbinary
jcm
jcm

Reputation: 5659

Are these two binary logic expressions equal?

Given the question:

Write C expressions, in terms of variable x, for the following values. Your code should work for any word size w ≥ 8. For reference, we show the result of evaluating the expressions for x = 0x87654321, with w = 32.

B. All but the least significant byte of x complemented, with the least significant byte left unchanged. [0x789ABC21].

I came up with the solution:

(~x)^0xFF but the provided solution was: x ^ ~0xFF - are these equivalent expressions?

Upvotes: 2

Views: 741

Answers (3)

chqrlie
chqrlie

Reputation: 144750

Actually, there are not always equivalent, and your solution is far better:

(~x) ^ 0xFF complements x for all integer types and restores the low order byte.

x ^ ~0xFF is more tricky: ~0xFF will be computed as an int and then extended to the type of x before xoring. It may not work as expected if x is an integer type larger than int and the representation of int is sign+magnitude.

Indeed, the following code:

unsigned long x = 12345678;
x = x ^ ~0xFF;

Produces the following warning with clang -Weverything:

xxo.c:4:13: warning: implicit conversion changes signedness: 'int' to 'unsigned long' [-Wsign-conversion]
x = x ^ ~0xFF;
      ~ ^~~~~

Furthermore, note that neither expression is correct if the size of w is smaller than the size of int. Say for example that int is 32 bits and short is 16 bits:

unsigned short x = 0x4321;

The expected result should be 0xBC21, but:

x ^ ~0xFF  ->  0x4321 ^ 0xFFFFFF00  ->  0xFFFFBC21

and

(~x) ^ 0xFF  ->  0xFFFFBCEF ^ 0xFF  ->  0xFFFFBC21

To fix this issue, the expression should be cast to the type of x, preferably an unsigned type.

Finally, note that you can shorten the expression this way: 0xFF ^ ~x

Upvotes: 1

Aviv
Aviv

Reputation: 424

  1. XOR is a symmetric binary operator. Therefore where lies the ~ is irrelevant.
  2. just a thinking pattern: take a sampled bit from x and track down its value: on our left expression every 1-bit becomes due to negation 0-bit and then 1-bit again after XORing with 0xFF constant. 0-bit for same reasons becomes 0-bits. on our right expression we happen to XOR x with 0 value in each bit. from XOR binary table it is easy to see that each bit in x will remain the same: 1-bits stays the same and 0-bits stays the same. And there you have it, expressions yield the same results. long road for simple answer...

Upvotes: 0

Hans Lepoeter
Hans Lepoeter

Reputation: 149

Yes.

And you can leave out (). ~ takes precedence over ^ ... and even if it didnt ( like ~(x^0xFF) ) you would still get the same result ...

Ain't that funny ?

Upvotes: 0

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