c++11 评估顺序(未定义行为)

Question

vec[ival++] <= vec[ival]

此表达式具有未定义的行为，因为运算符 (<=) 操作数的求值顺序未定义。

我们如何重写该表达式以避免未定义的行为？ 我找到了一个似乎有效的答案:

vec[ival] <= vec[ival + 1]

如果那是避免未定义行为的正确方法，为什么用这种方式重写它可以避免未定义行为？

添加任何关于如何修复该表达式的引用会很棒。

Answer 1

是的，您的第一个示例具有未定义的行为，因为我们对内存位置进行了无序的修改和访问，这是未定义的行为。这包含在 C++ 标准草案 [intro.execution]p10 中。 (强调我的):

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 memory location ([intro.memory]) is unsequenced relative to either another side effect on the same memory location or a value computation using the value of any object in the same memory location, and they are not potentially concurrent ([intro.multithread]), the behavior is undefined. [ Note: The next subclause imposes similar, but more complex restrictions on potentially concurrent computations. — end note  ] [ Example:

void g(int i) {
  i = 7, i++, i++;              // i becomes 9

  i = i++ + 1;                  // the value of i is incremented
  i = i++ + i;                  // the behavior is undefined
  i = i + 1;                    // the value of i is incremented
}

— end example  ]

如果我们查看涵盖 <= 的关系运算符部分[expr.rel]它没有指定评估顺序，所以我们被 intro.exection 涵盖了因此我们有未定义的行为。

具有未指定的评估顺序对于未定义的行为是不够的，如 Order of evaluation of assignment statement in C++ 中的示例演示。

您的第二个示例避免了未定义的行为，因为您没有向 ival 引入副作用，你只是在读取内存位置两次。

我相信这是解决问题的合理方法，它具有可读性并且不足为奇。另一种方法可以包括引入第二个变量，例如index和 prev_index .给定这么小的代码片段，很难得出快速规则。