我正在阅读 C++11 标准的 n3290 草案(尽可能接近实际的标准文本)，我注意到 i = i++ + 1; 会产生未定义的行为。我以前见过类似的问题，但他们是根据旧标准(序列点)回答的。新标准引入了表达式和子表达式执行之间的排序之前/之后关系的概念。

1.9 13 Sequenced before is an asymmetric, transitive, pair-wise relation between evaluations executed by a single thread (1.10), which induces a partial order among those evaluations. Given any two evaluations A and B, if A is sequenced before B, then the execution of A shall precede the execution of B. If A is not sequenced before B and B is not sequenced before A, then A and B are unsequenced. [ Note: The execution of unsequenced evaluations can overlap. —end note ] Evaluations A and B are indeterminately sequenced when either A is sequenced before B or B is sequenced before A, but it is unspecified which. [ Note: Indeterminately sequenced evaluations cannot overlap, but either could be executed first. —end note ]

1.9 14 Every value computation and side effect associated with a full-expression is sequenced before every value computation and side effect associated with the next full-expression to be evaluated.

1.9 15 Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced. [ Note: In an expression that is evaluated more than once during the execution of a program, unsequenced and indeterminately sequenced evaluations of its subexpressions need not be performed consistently in different evaluations. —end note ] The value computations of the operands of an operator are sequenced before the value computation of the result of the operator. If a side effect on a scalar object is unsequenced relative to either anotherside effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined.

[ Example:
void f(int, int);
void g(int i, int* v) {
i = v[i++]; // the behavior is undefined
i = 7, i++, i++; // i becomes 9
i = i++ + 1; // the behavior is undefined
i = i + 1; // the value of i is incremented
f(i = -1, i = -1); // the behavior is undefined
}
—end example ]

按照我的理解，它是这样工作的:

• operator= 有两个操作数表达式:引用 i 和 i++ + 1，两者都彼此无序。第二个对 i 有副作用，但在我看来第一个没有副作用或用于值计算(或者是引用采用“使用相同标量的值的值计算object”？它实际上是否依赖于存储在 i 中的值？不这么认为)，所以它不是未定义的行为；
• operator= 执行顺序在两个操作数评估之后进行。它对 i 有副作用，但它对两个操作数的引用顺序良好，因此不是未定义的行为；
• i++ + 1 显然是定义的行为。

我在这里有什么不对吗？还是因为其他原因，这条线未定义的行为？

PS。标准实际上是说

The value computations of the operands of an operator are sequenced before the value computation of the result of the operator.

，它根本没有提到这种情况下的副作用。然而，排序关系仅在表达式评估之间定义，并且评估=值计算+副作用。所以要么我必须假设这个草案在这里不一致，或者假设在这一行中他们意味着评估而不是值(value)计算。还是我错了？

编辑:

我想我会在这里回答自己，但这就是我困惑的原因:

5 1 An expression is a sequence of operators and operands that specifies a computation. An expression can result in a value and can cause side effects.

所以运算符的操作数本身不是子表达式。因此，仅对整个 i = i++ + 1; 的值计算进行了排序，标准中没有提及副作用排序。这就是它未定义的原因。

请注意，如果例如。 operator= 对给定类型进行了重载(因此这将是一个隐式函数调用)它不会是未定义的行为，对吧？

最佳答案

这是“未定义的行为”，而不是“未指定的”。未定义表示允许机器做任何事情，包括输出空程序、随机终止或爆炸。当然，当移植到另一个平台时，一个出乎意料的值是更可能的结果。

未定义的行为适用于两个副作用适用于相同标量而没有相对于彼此排序的任何情况。在这种情况下，副作用恰好是相同的(两者都从表达式之前的原始值增加 i)，但是按照标准的字母，它们结合起来产生 UB。

副作用是无序的，因为除了 、、?:、|| 和 &&，运算符不使用 C++11 §5.15/2 等术语定义排序规则:

If the second expression is evaluated, every value computation and side effect associated with the first expression is sequenced before every value computation and side effect associated with the second expression.

赋值运算符确实定义了一个特殊的排序规则，§5.17/1:

In all cases, the assignment is sequenced after the value computation of the right and left operands, and before the value computation of the assignment expression.

这对 i = i++ + 1 没有帮助，因为 i++ 的副作用不是任何值计算的一部分。

关于c++ - 为什么 C++11 中的 `i = i++ + 1` 行为未定义？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/10778627/