移位操作数在C ++ 17中排序 [英] Shift operands sequenced in C++17

问题描述

我读了C ++ 17 Standard $ 8.5.7.4:

I read in the C++17 Standard $8.5.7.4:

表达式E1在表达式E2之前排序.

The expression E1 is sequenced before the expression E2.

对于移位运算符.

cppreference 规则19表示:

In a shift operator expression E1<<E2 and E1>>E2, every value
computation and side-effect of E1 is sequenced before every value
computation and side effect of E2

但是当我尝试使用gcc 7.3.0或clang 6.0.0编译以下代码时

But when I try to compile the following code with gcc 7.3.0 or clang 6.0.0

#include <iostream>
using namespace std;

int main() {
    int i = 5;
    cout << (i++ << i) << endl;
    return 0;
}

我收到以下gcc警告:

I get the following gcc warning:

../src/Cpp_shift.cpp: In function ‘int main()’:
../src/Cpp_shift.cpp:6:12: warning: operation on ‘i’ may be undefined [-Wsequence-point]
  cout << (i++ << i) << endl;
           ~^~

叮当警告是:

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

我使用以下命令进行编译:

I used the following commands to compile:

g++ -std=c++17 ../src/Cpp_shift.cpp -o Cpp_shift -Wall
clang++ -std=c++17 ../src/Cpp_shift.cpp -o Cpp_shift -Wall

在两种情况下(5 * 2 ^ 6)我都得到了预期的320输出

I get the expected 320 as output in both cases ( 5 * 2 ^ 6 )

有人可以解释为什么我收到此警告吗?我有事吗我还阅读了与此相关的问题，但确实如此不回答我的问题.

Can someone explain why I get this warning? Did I overlook something? I also read this related question, but it does not answer my question.

所有其他变体++i << i，i << ++i和i << i++会产生相同的警告.

edit: all other variants ++i << i, i << ++i and i << i++ result in the same warning.

edit2:(i << ++i)对于clang(正确)导致320，对于gcc(错误)导致384.如果++位于E2，似乎gcc给出了错误的结果，(i << i++)也给出了错误的结果.

edit2: (i << ++i) results in 320 for clang (correct) and 384 for gcc (incorrect). It seems that gcc gives a wrong result if the ++ is at E2, (i << i++) also gives a wrong result.

推荐答案

标准清楚地说明了移位运算符操作数的求值顺序.

Standard is clear about the order of evaluation of the operands of the shift operator.

n4659-§8.8(p4):

n4659 - §8.8 (p4):

表达式E1在表达式E2之前排序.

The expression E1 is sequenced before the expression E2.

表达式i++ << i中没有未定义的行为，定义明确.它是 Clang 和

There is no undefined behavior in the expression i++ << i, it is well defined. It is a bug in Clang and GCC both.

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