重复使用前缀++运算符时出现未定义的行为 [英] Undefined behaviour in repeated use of prefix ++ operator

问题描述

我阅读了有关未定义行为的答案，其中看到以下语句：

I read this answer about undefined behaviour, where I saw following statement:

++++++i;     // UB, parsed as (++(++(++i)))

我不认为这是不确定的行为。我怀疑，这真的是C ++中的UB吗？如果是，那么如何？

I don't think it is undefined behaviour. I have a doubt, Is it really UB in C++? If yes, then How?

此外，我编写了程序并使用 g ++ prog.cpp -Wall -Wextra -std = gnu ++ 1z进行了编译-pedantic 命令，它工作正常，没有任何警告。

Also, I made program and compiled using g++ prog.cpp -Wall -Wextra -std=gnu++1z -pedantic command, it's working fine without any warning. It's give an expected output.

#include <iostream>
using namespace std;

int main()
{
    int i = 0;
    cout<<++++++i<<endl;
}

推荐答案

在C ++ 03中是未定义的行为。在C ++ 11中不是。各种预增量之间没有序列点。如果 i 是用户定义的类型，则它将是定义良好的行为，因为这样会进行函数调用（一个序列点）。

In C++03 it is undefined behavior. In C++11 it is not. There is no sequence point between the various pre-increments. If i was a user-defined type, it would be well-defined behavior because then there would be a function call (a sequence point).

在C ++ 11中，序列点的概念被之前/之后的序列代替。缺陷637（ http：//www.open-std。 org / jtc1 / sc22 / wg21 / docs / cwg_defects.html＃637 ）提供了一个示例，该示例以前未定义的构造变得定义明确（ i = ++ i +1 ）。

In C++11, the idea of sequence points was replaced with sequenced before/sequenced after. Defect 637 (http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_defects.html#637) provides an example of a previously undefined construct becoming well-defined (i = ++i + 1).

要了解为什么它不是未定义的行为，让我们看一下需要的部分。 ++ i 等效于 i = i + 1 （ i 仅被评估一次）。此外，如果我们用 inc 代替 i = i + 1 ，则 ++（i = i + 1）变为 inc = inc + 1 。

To understand why it's not undefined behavior, let's look at the pieces we need. ++i is equivalent to i = i + 1 (except i is evaluated only once). Further if we substitute i = i + 1 with inc, ++(i = i + 1) becomes inc = inc + 1.

[expr.ass]状态：

[expr.ass] states:

在所有情况下，赋值在左右操作数的值计算之后和赋值表达式的值计算之前被排序。

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.

因此， i = i + 1 在 inc 的值计算之前进行排序；但是，在 inc 的值计算之后，后， inc = inc + 1 中的赋值被排序。因为分配是顺序的，所以没有未定义的行为。

Thus the assignment in i = i + 1 is sequenced before value computation of inc; however, the assignment in inc = inc + 1 is sequenced after value computation of inc. There is no undefined behavior because the assignments are sequenced.

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