Cool Trick

Ah. Yet another iteration of the Swap Without A Temporary game.

The problems:

* Creating a temporary value on the stack is "free" anyways.
* This introduces numerous data dependencies, leading to pipeline stalls and reduced performance.
* Swaps do not take a significant amount of time in most code.
* Readability is important.
* If this were the best way to do it, why would your compiler's implementation of std::swap not do it this way?

Is there any sequence point problem in that?

x ^= y ^= x ^= y;

I actually got asked that question at an interview once. It's so usful to know

Quote:

Original post by rpg_code_master x ^= y ^= x ^= y; I actually got asked that question at an interview once. It's so usful to know

Above code has undefined behaviour. Can anyone tell me about my original code. Is it also undefined?

*bump*

Is the leftmost y in the equation's right hand side evaluated before or after the assignment of y? The standard does not say, so yes, it's undefined.

edit: And to demonstrate that, here's a test case.

Output from MSVC8 (both debug and release).

a = 42, b = 10a = 42, b = 42

Output from G++ (g++ test.cpp).

a = 42, b = 10a = 10, b = 42

Thanks Bro

Quote:

Original post by Lerner Quote: Original post by rpg_code_master x ^= y ^= x ^= y; I actually got asked that question at an interview once. It's so usful to know Above code has undefined behaviour. Can anyone tell me about my original code. Is it also undefined?

You are incorrect. The original code exhibits undefined behavior. This code's behavior is defined. The evaluation will proceed from the right hand side to the left.

Both are undefined. The OPs because of undefined order of evaluation of parameters and the xor swap because scalar objects may not have their values modified more than once by the evaluation of an expression without intervening sequence points (C++ Standard, Final Draft, Section 5, Paragraph 4).

Σnigma