简化此表达式

Question

Let a , b be positive integers with different values. Is there any way to simplify these expressions:

bool foo(unsigned a, unsigned b)
{
    if (a % 2 == 0)
      return (b % 2) ^ (a < b); // Should I write "!=" instead of "^" ?
    else      
      return ! ( (b % 2) ^ (a < b) ); // Should I write "(b % 2) == (a < b)"? 
}

I am interpreting the returned value as a boolean.

Answer 1

How is it different from

 (a%2)^(b%2)^(a<b)

which in turn is

 ((a^b)&1)^(a<b)

or, indeed

 ((a ^ b) & 1) != (a < b)

Edited to add: Thinking about it some more, this is just the xor of the first and last bits of (ab) (if you use 2's complement), so there is probably a machine-specific ASM sequence which is faster, involving a rotate instruction.

Answer 2

As a rule of thumb, don't mix operators of different operator families. You are mixing relational/boolean operators with bitwise operators, and regular arithmetic.

This is what I think you are trying to do, I'm not sure, since I don't understand the purpose of your code: it is neither readable nor self-explaining.

bool result;
bool a_is_even = (a % 2) == 0;
bool b_is_even = (b % 2) == 0;

if (a_is_even == b_is_even) // both even or both odd
  result = a < b;
else
  result = a >= b;

return result;

Answer 3

If you are returning a truth value, a boolean, then your proposed changes do not change the semantics of the code. That's because bitwise XOR, when used in a truth context, is the same as != .

In my view your proposed changes make the code much easier to understand. Quite why the author thought bitwise XOR would be appropriate eludes me. I guess some people think that sort of coding is clever. I don't.

If you want to know the relative performance of the two versions, write a program and time the difference. I'd be surprised if you could measure any difference between them. And I'd be equally surprised if these lines of code were your performance bottleneck.

Answer 4

I program in C# but I'd think about something like this:

return (a % 2 == 0) && ((b % 2) ^ (a < b))

Considering from you comments that '^' is equivalent to '=='

Answer 5

Since there is not much information around this issue, try this:

int temp = (b % 2) ^ (a < b);
if (a % 2 == 0)
  return temp;
else      
  return !temp;  

This should be less code if the compiler hasn't optimized it already.