通过XOR在单行中交换整数。是否真的允许在c ++ 11？ [英] Swap integers via XOR in single line. Is it really allowed in c++11?

问题描述

我还是不清楚是否在C ++ 11中有效的表达式 x ^ = y ^ = x ^ = y;
在此线程），否则会导致未定义的行为？

I still could not clearly understand whether the expression x ^= y ^= x ^= y; valid in C++11 (as they say in this thread) or it leads to undefined behavior?

链接给出的原因似乎令人信服，但clang引发了一个警告：

The reasons given by the link seem convincing, but clang throws a warning:

警告：无法修改并访问'x'[-Wunsequenced]

warning: unsequenced modification and access to 'x' [-Wunsequenced]

此外，如果两个版本：

x ^= y ^= x ^= y; // (1) 
x = x ^ (y = y ^ (x = (x ^ y))); // (2)

被认为是等效的（并且在C ++ 11中定义良好）给出不同的结果（第一，第二个）？

considered equivalent (and well-defined in C++11), why it gives different results (first, second)?

此外，应该注意，gcc给出了

Additionally, it should be noted that the gcc gives a warning about sequence point only on the second version of code.

推荐答案

关于序列点的讨论，请参阅下面的例子：crooked.com/a/ff6bb101c3b55ff4\">

赋值运算符（=）和复合赋值运算符全部由
组从右到左。 [..]

形式 E1 op = E2 的表达式的行为相当于 E1 = E1 op E2 ，除了 E1 是
只评估一次。

The assignment operator (=) and the compound assignment operators all group right-to-left. [..]
The behavior of an expression of the form E1 op = E2 is equivalent to E1 = E1 op E2 except that E1 is evaluated only once.

因此，您的代码等同于

x = x ^ (y ^= (x ^= y)));

...在 x = x ...中只评估一次。
不幸的是，对于xor的，对操作数的评估是无效的。也就是说

... with x evaluated only once in x = x .... Unfortunately, for xor's, the evaluation of the operands is unsequenced. I.e.

除非另有说明，否则对个别运算符
的操作数和各个表达式的子表达式的计算结果无效。

Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced.

适用。但现在我们有一个问题：

applies. But now we have a problem:

   x = x ^ (y ^= (x ^= y)));
//     *          ******
//     |            |
//     |            Side effect
//     Value computation

价值计算 x 对于最左边的两个 x ）和副作用不相互排序，因此我们诱导UB： / p>



The value computation (which is implied in the singular evaluation of x for the two leftmost x) and the side effect are unsequenced wrt each other, hence we induce UB:

如果对标量对象的副作用不是相对于
的顺序，对同一标量对象的另一个副作用或值计算
使用相同标量对象的值，行为是未定义的。

If a side effect on a scalar object is unsequenced relative to either another side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined.

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