Why don't people use xor swaps? [closed]

问题:

I read on a site that using xor swaps is fast because it doesn't use a temporary variable. Here's an example:

#include <stdio.h>

int main(void)
{        
    int a=234,b=789;
    b=b^a;
    a=b^a;
    b=b^a;
    printf("a=%d,b=%d",a,b);
    return 0;
}

Why don't people use this technique in real life code? Is it just poor style? Is there something not well defined about it? Is it an optimisation that my compiler might produce from more clear code, automatically?

回答1:

1. Because readability is preferred over performance.

2. Because tmp = a; a = b; b = tmp; is not that slow.

3. Because the compiler will optimize it anyway.

4. Because it works for integers only. What if you want to swap floating-point numbers? Strings? Custom objects? etc.

回答2:

All answers are already there consider it just an addition-

->if both values goes for same memory address-result will be zero

->compilers can optimize away the temporary variable in the naive swap

->modern CPUs strive to execute instructions in parallel via instruction pipelines but with XOR technique is considerably slower than using a temporary variable to do swapping because each operation depends on the result of previous

->x+Y may go for integer overflow

回答3:

1. While there is no explicit temporary variable, the results are actually stored in an implicit temp variable before being written to the register.

2. With xor swap, you need to ensure that the variables being swapped aren't same. Else both shall be evaluated to 0.

回答4:

The performance gain is typically so small that the cost to "understandable code" is higher than the speed benefit obtained.