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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);} 