湖南汽车软件网站建设,免费网站空间 asp.net,wordpress 更换首页,wordpress和微信公众号互通接触过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#xff0c;特别是涉及到struct#xff08;结构体#xff09;和union#xff08;联合体#xff09;时。今天我们通过详细例子来说明下struct和union的对齐规则#xff0c;以便了解各种struct和… 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n1)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cda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bcd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);}
