C++的struct类型的内存问题

在用struct定义一个结构体的时候，突发想象：struct变量中的不同类型是怎么样分配内存的？

于是，用VC做了个小实验

struct中定义了char、int、float等类型数据

先来看看只有一个char类型数据，一个int类型数据，一个float类型数据，一个double类型数据的struct类型

struct Structure1 {  char c1;  int i;  float f;  double d;};#include <iostream>using namespace std;//typedef struct Structure1 struction;int main() {    struct Structure1 s1, s2;    int size = sizeof s1;  cout<<"Size of char is "<<sizeof(char)<<endl;  cout<<"Size of int is "<<sizeof(int)<<endl;  cout<<"Size of float is "<<sizeof(float)<<endl;  cout<<"Size of double is "<<sizeof(double)<<endl;  cout<<"Size of struct is "<<size<<endl;  s1.c1 = 'a'; // Select an element using a '.'  s1.i = 1;  s1.f = 3.14;  s1.d = 0.00093;  cout<<"Size of struct is "<<size<<endl;  cout<<"Address of s1 is "<<&s1<<endl;  if((&s1) == (struct Structure1*)((void*)&(s1.c1)))  {  cout<<"s1 share the same address with s1.c1!"<<endl;  }  else  {  cout<<"s1 didn't share the same address with s1.c1!"<<endl;  }  cout<<"Address of s1.c1 is "<<&(s1.c1)<<endl;  cout<<"Address of s1.i is "<<&(s1.i)<<endl;  cout<<"Address of s1.f is "<<&(s1.f)<<endl;  cout<<"Address of s1.d is "<<&(s1.d)<<endl;  cout<<"Size of s1.d is "<<sizeof s1.d<<endl;  cout<<"Size of s1.c1 is "<<sizeof s1.c1<<endl;  s2.c1 = 'a';  s2.i = 1;  s2.f = 3.14;  s2.d = 0.00093;} 

运行结果

看到的是本来char型只有一个字节，但是从int地址和s1的起始地址来看，char型应该是占据了四个字节。这其实是一个内存对齐的问题，主要跟操作系统分配内存的单元大小有关。

所以下面，我就多添加几个char型

struct Structure1 {  char c1;  int i;  float f;    char c2;//让c2在定义的时候不跟在c1的后面，也就是想看看有没有内存对齐的问题  char c3;  char c4;  double d;};#include <iostream>using namespace std;int main() {    struct Structure1 s1, s2;    int size = sizeof s1;  cout<<"Size of char is "<<sizeof(char)<<endl;  cout<<"Size of int is "<<sizeof(int)<<endl;  cout<<"Size of float is "<<sizeof(float)<<endl;  cout<<"Size of double is "<<sizeof(double)<<endl;  cout<<"Size of struct is "<<size<<endl;  s1.c1 = 'a'; //   s1.i = 1;  s1.f = 3.14;  s1.d = 0.00093;  cout<<"Size of struct is "<<size<<endl;  cout<<"Address of s1 is "<<&s1<<endl;  if((&s1) == (struct Structure1*)((void*)&(s1.c1)))//判断有没有c1的地址是和struct变量的起始地址是一样的  {  cout<<"s1 share the same address with s1.c1!"<<endl;  }  else  {  cout<<"s1 didn't share the same address with s1.c1!"<<endl;  }  cout<<"Address of s1.c1 is "<<&(s1.c1)<<endl;  cout<<"Address of s1.c2 is "<<&(s1.c2)<<endl;  cout<<"Address of s1.c3 is "<<&(s1.c3)<<endl;  cout<<"Address of s1.i is "<<&(s1.i)<<endl;  cout<<"Address of s1.f is "<<&(s1.f)<<endl;  cout<<"Address of s1.d is "<<&(s1.d)<<endl;  cout<<"Size of s1.d is "<<sizeof s1.d<<endl;  cout<<"Size of s1.c1 is "<<sizeof s1.c1<<endl;  cout<<"Size of s1.c2 is "<<sizeof s1.c2<<endl;  cout<<"Size of s1.c3 is "<<sizeof s1.c3<<endl;} 

运行的结果如下：

可以看出虽然c2、c3并不是和c1在一起定义的，但是分配内存的时候还是没有另外分配四个字节，而是利用c1后面没有分配完字节给c2、c3

这里只是简单讨论了char型，当然大家还可以去思考其他诸如class等复杂的类型并验证