详解下c语言中struct和union的对齐规则

    接触过c语言的同学应该都知道字节对齐。有些时候我们很容易弄错字节对齐的方式，特别是涉及到struct（结构体）和union（联合体）时。今天我们通过详细例子来说明下struct和union的对齐规则，以便了解各种struct和union所占字节具体计算方式。

一、基础环境信息

    我本地计算机系统各类基础类型所占字节如下：

int main() {printf("the sizeof char      is %d\n",sizeof(char));printf("the sizeof int       is %d\n",sizeof(int));printf("the sizeof unsigned  is %d\n",sizeof(unsigned));printf("the sizeof short     is %d\n",sizeof(short));printf("the sizeof long      is %d\n",sizeof(long));printf("the sizeof float     is %d\n",sizeof(float));printf("the sizeof double    is %d\n",sizeof(double));printf("the sizeof longlong  is %d\n",sizeof(long long));}

二、结构体字节对齐

2.1，结构体对齐规则

    结构体中寻找所有成员中占字节数最大成员，其余成员根据占字节数最大成员拼凑或者插入空位构成n个（n>=1)最大成员字节。

2.2，实例

    上图例子实测：

int main() {struct stu{char a;int b;short c;char d;};struct  stu s;printf("the sizeof s is %d\n", sizeof s);printf("the address a is %x\n", &s.a);printf("the address b is %x\n", &s.b);printf("the address c is %x\n", &s.c);printf("the address d is %x\n", &s.d);
}

    对于如下结构体：最大类型为b，占4字节，c，d，a拼凑占4字节，总共8字节

struct stu{int b;short c;char d;char a;
};

int main() {struct stu{int b;short c;char d;char a;};struct  stu s;printf("the sizeof s is %d\n", sizeof s);printf("the address b is %x\n", &s.b);printf("the address c is %x\n", &s.c);printf("the address d is %x\n", &s.d);printf("the address a is %x\n", &s.a);
}

 

    对于如下结构体： f为最大类型，占8字节，成员e需要以8字节对齐，所以stu1占16字节；成员b，c，d拼凑占8字节；成员a分配8字节进行对齐；总共占32字节

struct stu{int b;short c;char d;struct stu1{char e;double f;} stu1;char a;
};

int main() {struct stu{int b;short c;char d;struct stu1{char e;double f;} stu1;char a;};struct  stu s;printf("the sizeof s     is %d\n", sizeof s);printf("the address b    is %x\n", &s.b);printf("the address c    is %x\n", &s.c);printf("the address d    is %x\n", &s.d);printf("the address stu1 is %x\n", &s.stu1);printf("the address e    is %x\n", &s.stu1.e);printf("the address f    is %x\n", &s.stu1.f);printf("the address a    is %x\n", &s.a);
}

三、联合体字节对齐

3.1，联合体对齐规则

    联合体字节对齐计算非常简单，为所有成员中所占字节最大成员的字节数。

3.2，实例

    上图例子实测：

int main() {union stu{char a;int b;short c;char d;};union  stu s;printf("the sizeof s     is %d\n", sizeof s);printf("the address b    is %x\n", &s.b);printf("the address c    is %x\n", &s.c);printf("the address d    is %x\n", &s.d);printf("the address a    is %x\n", &s.a);s.b = 0b00000011000000110000001100000011;printf("the value of b    is %d\n", s.b);printf("the value of c    is %d\n", s.c);printf("the value of d    is %d\n", s.d);printf("the value of a    is %d\n", s.a);
}

 

    对于如下联合体：stu中成员stu1按照struct对齐规则占8字节，所以联合体stu占8字节。

union stu{char a;int b;short c;struct stu1{char e;int f;}stu1;char d;
};

int main() {union stu{char a;int b;short c;struct stu1{char e;int f;}stu1;char d;};union  stu s;printf("the sizeof s     is %d\n", sizeof s);printf("the address a    is %x\n", &s.a);printf("the address b    is %x\n", &s.b);printf("the address c    is %x\n", &s.c);printf("the address e    is %x\n", &s.stu1.e);printf("the address f    is %x\n", &s.stu1.f);printf("the address d    is %x\n", &s.d);}