c++ - 打包成员的引用地址不等于打包成员的地址？

Question

我用 g++ 编译了这个程序。它打印出“不平等”。
如果我不使用“packed”属性，它会打印“equal”。
我不知道“打包”属性如何导致差异。

顺便说一句，由 clang++ 构建的可执行文件打印“相等”。

#include <iostream>

struct __attribute__ ((packed)) Packed {
  char a;
  int b;
  int c;
  char d;
};

void test(const int &i, int *ptr) {
  std::cout << ((&i == ptr) ? "equal" : "unequal") << std::endl;
}

int main () {
  Packed p;
  p.c = 1;
  test(p.c, &p.c);
  return 0;
}

标签： c++

Answer 1

When compiling with GCC or CLANG, there is a warning saying that taking address of packed member of 'Packed' may result in an unaligned pointer value which means that behavior of your code is unspecified. That's why you have different outputs when compiling with GCC and CLANG.

What does the __attribute__ ((packed)) mean?

The packed variable attribute specifies that a structure field has the smallest possible alignment. That is, one byte for a variable field, and one bit for a bitfield, unless you specify a larger value with the aligned attribute.

Why not to take the address of a packed member?

Taking the address of a packed member is dangerous since the reduced alignment of the pointee is lost. This can lead to memory alignment faults in some architectures if the pointer value is dereferenced.

References:

• Warn when taking address of packed member
• __attribute__((packed)) variable attribute