Do bitwise operators (other than shifts) make any mathematical sense in base-10?

Question

According to wiki shifts can be used to calculate powers of 2:

A left arithmetic shift by n is equivalent to multiplying by 2^n (provided the value does not overflow), while a right arithmetic shift by n of a two's complement value is equivalent to dividing by 2^n and rounding toward negative infinity.

I was always wondering if any other bitwise operators (~,|,&,^) make any mathematical sense when applied to base-10? I understand how they work, but do results of such operations can be used to calculate anything useful in decimal world?

Answer 1

 "yep base-10 is what I mean" 

In that case, yes, they can be extended to base-10 in several ways, though they aren't nearly as useful as in binary.

One idea is that &, |, etc. are the same as doing arithmetic mod-2 to the individual binary digits. If a and b are single binary-digits, then

 a & b = a * b (mod 2) a ^ b = a + b (mod 2)    ~a = 1-a   (mod 2) a | b = ~(~a & ~b) = 1 - (1-a)*(1-b) (mod 2) 

The equivalents in base-10 would be (note again these are applied per-digit, not to the whole number)

 a & b = a * b (mod 10) a ^ b = a + b (mod 10)    ~a = 9-a   (mod 10) a | b = ~(~a & ~b) = 9 - (9-a)*(9-b) (mod 10) 

The first three are useful when designing circuits which use BCD (~a being the 9's complement), such as non-graphing calculators, though we just use * and + rather than & and ^ when writing the equations. The first is also apparently used in some old ciphers.

Answer 2

A fun trick to swap two integers without a temporary variable is by using bitwise XOR:

void swap(int &a, int &b) {    a = a ^ b;    b = b ^ a; //b now = a    a = a ^ b; //knocks out the original a } 

This works because XOR is a commutative so a ^ b ^ b = a.