使用C语言编写一个关于结构的示例程序

The structure is a collection of different datatype variables, grouped together under a single name Syntax.

Declaration and initialization of structures

The general form of structure declaration is as follows −

datatype member1;
struct tagname{
   datatype member2;
   datatype member n;
};

在这里，struct - 关键字

         tagname - 指定结构的名称

         member1, member2 - 指定构成结构的数据项。

示例

struct book{
   int pages;
   char author [30];
   float price;
};

结构变量

声明结构变量有三种方式。它们如下所示 −

1) struct book{
   int pages;
   char author[30];
   float price;
}b;
2) struct{
   int pages;
   char author[30];
   float price;
}b;
3) struct book{
   int pages;
   char author[30];
   float price;
};
struct book b;

结构的初始化和访问

• 成员和结构变量之间的链接是通过成员运算符（或者点运算符）建立的。

• 初始化可以通过以下方式进行：

方法1

struct book{
   int pages;
   char author[30];
   float price;
} b = {100, “balu", 325.75};

Method 2

struct book{
   int pages;
   char author[30];
   float price;
};
struct book b = {100, “balu", 325.75};

Method 3 (using member operator)

struct book{
   int pages;
   char author[30];
   float price;
} ;
struct book b;
   b. pages = 100;
   strcpy (b.author, “balu");
   b.price = 325.75;

方法四（使用scanf函数）

struct book{
   int pages;
   char author[30];
   float price;
} ;
struct book b;
   scanf (“%d", &b.pages);
   scanf (“%s", b.author);
   scanf (“%f", &b. price);

We can print the contents of either of the above structures in the main method as shown below −

main ( ){
   struct book b;
   clrscr ( );
   printf ( "enter no of pages, author, price of book");
   scanf ("%d%s%f", &b.pages, b.author, &b.price);
   printf("Details of book are");
   printf("pages =%d, author = %s, price = %f", b.pages, b.author, b.price);
   getch();
}

Example

Following is another example of structures −

 Live Demo

#include<stdio.h>
struct aaa{
   struct aaa *prev;
   int i;
   struct aaa *next;
};
main(){
   struct aaa abc,def,ghi,jkl;
   int x=100;
   abc.i=0;
   abc.prev=&jkl;
   abc.next=&def;
   def.i=1;
   def.prev=&abc;
   def.next=&ghi;
   ghi.i=2;ghi.prev=&def;
   ghi.next=&jkl;
   jkl.i=3;
   jkl.prev=&ghi;
   jkl.next=&abc;
   x=abc.next->next->prev->next->i;
   printf("%d",x);
}

输出

2

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