使用程序存储学生信息的C程序

编写一个C程序来存储使用Structure的学生信息。每个要存储的学生的信息是：

Each Student Record should have:
  Name
  Roll Number
  Age
  Total Marks

• 结构是C / C++中用户定义的数据类型。结构创建一个数据类型，该数据类型可用于将可能不同类型的项目分组为单个类型。
• ‘struct’关键字用于创建学生结构，如下所示：

  struct Student
  { 
     char* name; 
     int roll_number;
     int age;
     double total_marks;
  };
  

• 获取要存储其详细信息的学生数。为了简单起见，我们在这里招收5名学生。
• 创建一个Student结构变量以访问记录。在这里被视为“学生”
• 获取n个学生的数据，并通过点(。)运算符将其存储在学生的字段中
  句法：

  student[i].member = value;

• 存储所有数据后，使用点(。)运算符打印每个学生的记录并循环。

  句法：

  student[i].member;

下面是上述方法的实现：

// C Program to Store Information
// of Students Using Structure
  
#include 
#include 
#include 
  
// Create the student structure
struct Student {
    char* name;
    int roll_number;
    int age;
    double total_marks;
};
  
// Driver code
int main()
{
    int i = 0, n = 5;
  
    // Create the student's structure variable
    // with n Student's records
    struct Student student[n];
  
    // Get the students data
    student[0].roll_number = 1;
    student[0].name = "Geeks1";
    student[0].age = 12;
    student[0].total_marks = 78.50;
  
    student[1].roll_number = 5;
    student[1].name = "Geeks5";
    student[1].age = 10;
    student[1].total_marks = 56.84;
  
    student[2].roll_number = 2;
    student[2].name = "Geeks2";
    student[2].age = 11;
    student[2].total_marks = 87.94;
  
    student[3].roll_number = 4;
    student[3].name = "Geeks4";
    student[3].age = 12;
    student[3].total_marks = 89.78;
  
    student[4].roll_number = 3;
    student[4].name = "Geeks3";
    student[4].age = 13;
    student[4].total_marks = 78.55;
  
    // Print the Students information
    printf("Student Records:\n\n");
    for (i = 0; i < n; i++) {
        printf("\tName = %s\n", student[i].name);
        printf("\tRoll Number = %d\n", student[i].roll_number);
        printf("\tAge = %d\n", student[i].age);
        printf("\tTotal Marks = %0.2f\n\n", student[i].total_marks);
    }
  
    return 0;
}

输出：

Student Records:

    Name = Geeks1
    Roll Number = 1
    Age = 12
    Total Marks = 78.50

    Name = Geeks5
    Roll Number = 5
    Age = 10
    Total Marks = 56.84

    Name = Geeks2
    Roll Number = 2
    Age = 11
    Total Marks = 87.94

    Name = Geeks4
    Roll Number = 4
    Age = 12
    Total Marks = 89.78

    Name = Geeks3
    Roll Number = 3
    Age = 13
    Total Marks = 78.55

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