c language structure and union

Structure

Basic knowledge of structure

1. The difference between structure and array
   
2. Create a structure

struct student  //学生的数据类型
{
    
    
	char name[10];
	int age;

};

You can also create structure pointers and structure arrays

struct student* p; //结构体指针 ++跳过一整个结构体
struct student arr[10]; // 结构体数组 存放10个

3. Structure access
   
   Note:

#include<stdio.h>

struct student
{
    
    
	char name[20];
	int age;
}x,*p;

int main()
{
    
    
	
	if (p == &x)
	{
    
    
		printf("haha");
	}
	return 0;
}

Here
p is different from &x. p is a structure pointer but it is a null pointer, which is equivalent to the following behavior

int a;

&x is a structure pointer that points to x after the address of x is saved with a structure pointer.

4. The use of
   typedef typedef can simplify a structure variable name

typedef struct book
{
    
    
	char name[10];
	int price;
} book;


int main()
{
    
    
	book x[10] = {
    
     0 };
	return 0;
}

Now book is a structure type.

5. Direct access to structure

typedef struct book
{
    
    
	char name[10];
	int price;
} book;


int main()
{
    
    
	book x[10] = {
    
     0 };
	x[1].name[10] = "abcd";  //通过 点操作符访问
	return 0;
}

6. Indirect access via pointer

typedef struct
{
    
    
	char name[10];
	int price;
}book;

void fun(book* pc)
{
    
    
	pc->price =10;
	printf("%d",pc->price);
}


int main()
{
    
    
	 book x[10] = {
    
     0 };
	
	 fun(x);
	return 0;
}

Structure pointer

struct stu
{
    
    
	char name[10];
	int age;
};

int main()
{
    
    
	
	struct stu *p;


	return 0;
}

p is a pointer to a structure p+1 can jump to the next structure
*p+1 this is an illegal operation
*p and p *P is the entire structure p is a structure pointer

#include<stdio.h>

struct stu
{
    
    
	char name[10];
	int age;
}x;

int main()
{
    
    
	
	struct stu *p=&x;
	p->age = 10; // 可以这样 指针指向
	(*p).age=10; // 可以解引用得到 x
	return 0;
}

2. Structure pointer access to nested structure

#include<stdio.h>

struct stu
{
    
    
	int arr[2];
	int age;
};

struct ptu
{
    
    
	int a;
	int b;
	struct stu sr ;
}s;

int main()
{
    
    
	struct ptu* p = &s;
	p->sr.age;
	return 0;
}

Structure memory allocation

First of all, what will this code output?

struct S1
{
    
    
	char c1;
	int i;
	char c2;
};
int main()
{
    
    
printf("%d\n", sizeof(struct S1));
return 0;
}

Output 12 shouldn't be 1+4+1=6. It can be seen that the structure memory is stored differently.

12. 第一个成员在与结构体变量偏移量为0的地址处。
13. 其他成员变量要对齐到某个数字（对齐数）的整数倍的地址处。
对齐数 = 编译器默认的一个对齐数 与 该成员大小的较小值。

VS中默认的值为8
Linux中的默认值为4
14. 结构体总大小为最大对齐数（每个成员变量都有一个对齐数）的整数倍。
15. 如果嵌套了结构体的情况，嵌套的结构体对齐到自己的最大对齐数的整数倍                                                 
16. 结构体的整体大小就是所 有最大对齐数（含嵌套结构体的对齐数）的整数倍。

Green char blue useless memory red int

joint

Poke here to unite