C ++：一个空类的对象的大小是多少？ [英] C++: What is the size of an object of an empty class?
问题描述
我想知道什么是空类的对象的大小。它肯定可以不是为0字节，因为它应该可以像任何其他对象一样引用和指向它。
I was wondering what could be the size of an object of an empty class. It surely could not be 0 bytes since it should be possible to reference and point to it like any other object. But, how big is such an object?
I used this small program: c $ c> #include< iostream>
using namespace std;
class Empty {};
int main（）
{
Empty e;
cerr<< sizeof（e） endl;
return 0;
}
#include <iostream>
using namespace std;
class Empty {};
int main()
{
Empty e;
cerr << sizeof(e) << endl;
return 0;
}
Visual C ++和Cygwin-g ++编译器上的输出是 1个字节！这对我来说有点意外，因为我期望它是机器字的大小（32位或4字节）。
The output I got on both Visual C++ and Cygwin-g++ compilers was 1 byte! This was a little surprising to me since I was expecting it to be of the size of the machine word (32 bits or 4 bytes).
任何人都可以解释为什么 1字节的大小？ 为什么不 4个字节？这是依赖于编译器还是机器呢？此外，有人可以更清楚地说明为什么空类对象不会大小为0字节。
Can anyone explain why the size of 1 byte? Why not 4 bytes? Is this dependent on compiler or the machine too? Also, can someone give a more cogent reason for why an empty class object will not be of size 0 bytes?
推荐答案
引用 Bjarne Stroustrup的C ++风格和技巧常见问题，原因大小为非零是确保两个不同对象的地址将不同。尺寸可以是1，因为在这里对齐不重要，因为没有什么可以实际看。
Quoting Bjarne Stroustrup's C++ Style and Technique FAQ, the reason the size is non-zero is "To ensure that the addresses of two different objects will be different." And the size can be 1 because alignment doesn't matter here, as there is nothing to actually look at.
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