为什么类对象的大小基于成员的顺序不同？ [英] Why is the size of a class object different based on order of members?

问题描述

class CHaraICICCC
{
int i;
char c1;
int j;
char c2;
char c3;
char c4;
};

class CHaraIICCCC
{
int i;
int j;
char c1;
char c2;
char c3;
char c4;
};

void fun()
{
    CHaraICICCC eici;
    CHaraIICCCC eiicc;

    int icic = sizeof(eici); // -> output of icic is 16.
    int iicc = sizeof(eiicc); // -> output of icic is 12.
}

如果有人知道，请让我知道为什么喜欢这个。
感谢
Hara

If any one knows, Please let me know why like this. Thanks Hara

推荐答案

由于对齐。 x86编译器倾向于在4字节边界上对齐int类型（用于更快的内存访问），因此CHaraICICCC可能会被布局为：

Because of alignment. x86 compilers tend to align int types on 4 bytes boundary (for faster memory access) so CHaraICICCC would probably be laid out as this:

byte  0: \
byte  1: | <--- int i
byte  2: |
byte  3: /  
byte  4: <----- char c1
byte  5: \
byte  6: | <--- Padding (wasted bytes)
byte  7: / 
byte  8: \
byte  9: | <--- int j  
byte 10: |
byte 11: /
byte 12: <----- char c2
byte 13: <----- char c3
byte 14: <----- char c4

总共15个字节，而CHaraIICCCC将是：

for a total of 15 bytes, while CHaraIICCCC would be:

byte  0: \
byte  1: | <--- int i
byte  2: |
byte  3: /  
byte  4: \
byte  5: | <--- int j 
byte  6: |
byte  7: /
byte  8: <----- char c1
byte  9: <----- char c2
byte 10: <----- char c3
byte 11: <----- char c4

总共12个字节（没有字节浪费填充）。当然，这是很多编译器相关的，取决于你的编译选项。

for a total of 12 bytes (with no bytes wasted for padding). Of course, this is much compiler-related and dependant on your compilation options.

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