如何计算类和填充的大小 [英] How do I calculate size of class and padding

问题描述

#include "stdafx.h"
#include"iostream"

class test
{
public:
	int a ;//size of integer is 4  byte
	double b;//size of double is 8 byte
	char ch[8+1];///9 size of char array is 9 byte

};
int _tmain(int argc, _TCHAR* argv[])
{
	test obj;
	std::cout<<sizeof(obj.a)<<std::endl;
	std::cout<<sizeof(obj.b)<<std::endl;
	std::cout<<sizeof(obj.ch)<<std::endl;
	std::cout<<sizeof(test); //why size of class is 32 byte??
	
	
	int p;
	std::cin>>p;
	return 0;
}

输出：

4

8

9

32



我真的不明白为什么班级规模大小32字节。如果我只计算三个数据成员的大小，它只有21个字节。我确实理解使用了一些字节概念但我不明白字节填充是如何工作的。



任何人都可以告诉我这个。如果我改变了成员的顺序数据类的大小将如何变化以及为什么？



问候，

Joy

output:
4
8
9
32

I really do not understand why size of class is 32 byte. if I calculate size of three data members alone its just 21 byte only. I do understand that some byte concept is used but I do not understand how byte padding works.

Can anybody make me clear on this.And if I change order of member data how size of class will get changed and why?

Regards,
Joy

推荐答案

Padding。

Microsoft C ++编译器（正如Peter给你的链接所说）总是将类填充到一个合理大小的内存访问。

如果你取出了char数组，你会发现它是16个字节。

如果你把它减少到：

Padding.
The Microsoft C++ compiler (as the link Peter gave you says) always pads classes to a "sensible" size for memory access.
If you take out the char array, you will find it is 16 bytes.
If you cut it down to:

class test
    {
    public:
        int x;
        char c;
    };

你会发现它被填充到8个字节。

You will find it is padded to 8 bytes.

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