结构体内存对齐规则

1—————————————–

struct{    char b;    int a;    short c;}A;sizeof(A) = 12;b: 1*0 = 0a: 4*1 = 4c: 2*4 = 8 2-----

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struct A{    int a;    double b;    float c;};struct{    char e[2];    int f;    double g;    short h;    struct A i;}B;//对齐单位 8 个字节sizeof(B) = 48;//普通成员偏移量e: 2*0 = 0f: 4*1 = 4g: 8*1 = 8h: 2*8 = 16结构体起点坐标：   8*3 = 24//结构体成员偏移量a: 24 + 4*0 = 24b: 24 + 8*1 = 32c: 24 + 4*4 = 403------------------------

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struct A{    char e;    short f;};struct{    int a;    char b;    struct A c;    char d;}B;sizeof(B) = 12;a: 4*0 = 0;b: 1*4 = 4;结构体起点坐标：    2*3 = 6e: 6+1*0 = 6f: 6+2*1 = 8d: 1*10 = 10

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4-----------------struct A{    int e;    short f;};struct{    int a;    char b;    struct A c;    short d;}B;sizeof(B) = 20;a: 4*0 = 0b: 1*4 = 4结构体起点：    4*2 = 8e: 8 + 4*0 = 8f: 8 + 2*2 = 12 //结构体A所占的大小为该结构体成员内部最大元素的整数倍，不足补齐。d: 2*8 = 165-----------------

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#pragma pack(2)  //指定对齐单位为2个字节typedef struct{    int a;     char b;    short c;    char d;}A;

a:2*0 = 0
b: 1*4 = 4
c: 2*3 = 6
d: 1*8 = 8
sizeof(A) = 10;

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#pragma pack(2)typedef struct{    char a;     char b;    short c;    int d;}A;a: 1*0 = 0b: 1*1 = 1c: 2*1 = 2d: 2*2 = 4sizeof(A) = 8;