潜在问题在＆QUOT;交换两个变量值，而无需使用第三个变量＆QUOT; [英] Potential Problem in "Swapping values of two variables without using a third variable"

问题描述

我最近沿此方法来对交换两个变量的值，而无需使用第三变量

A ^ = B ^ = a ^ = B

但是，当我尝试了不同的编译器上面的code，我得到了不同的结果，有的给了正确的结果，有些却没有。

什么是可怕的错误与code？

解决方案

  

什么是可怕的错误与code？

是的！

A ^ = B ^ = A ^ = B 实际上调用未定义行为的C和C ++中，因为你正在试图更改在两个以上的序列点之间的一次。

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试着写（尽管并非万无一失）

  A ^ = B;
b ^ =一;
一个^ = B;
 

而不是 A ^ = B ^ = a ^ = B 。

P.S ：切勿试图交换两个变量的值，而无需使用第三个。始终使用第三个变量。

修改

由于 @caf 注意到 B ^ = a ^ = B 细纵然 ^ = 运营商是不确定的，因为的所有访问b 恩pression内被用来计算正在被存储在 b 终值，该行为被明确定义。

I recently came along this method for swapping the values of two variables without using a third variable.

a^=b^=a^=b

But when I tried the above code on different compilers, I got different results, some gave correct results, some didn't.

Is anything terribly wrong with the code?

解决方案

Is anything terribly wrong with the code?

Yes!

a^=b^=a^=b in fact invokes Undefined Behaviour in C and in C++ because you are trying to change the value of a more than once between two sequence points.

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Try writing (although not foolproof )

a ^= b;
b ^= a;
a ^= b;

instead of a^=b^=a^=b.

P.S : Never try to swap the values of two variables without using a third one. Always use a third variable.

EDIT :

As @caf noticed b^=a^=b is fine even though the order of evaluation of arguments of ^= operator is unspecified, since all the accesses of b within the expression are being used to compute the final value that is being stored in b, the behaviour is well defined.

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