c语言-结构体实例笔记

结构体实例

实例一览：

• 使用结构体存储学生的信息

  Store information of a student using structure

• 计算二者距离（以英寸英尺为单位）

  Add two distances (in inch-feet)

• 通过结构体传递给函数来计算两个复数相加

  Add two complex numbers by passing structures to a function

• 计算两个时间段之间的差

  Calculate the difference between two time periods

• 使用结构体存储10名学生的信息

  Store information of 10 students using structures

• 使用结构体存储n名学生的信息

  Store information of n students using structures

Store information of a student using structure

#include 
struct student {
     
    char name[50];
    int roll;
    float marks;
} s;

int main() {
     
    printf("Enter information:\n");
    printf("Enter name: ");
    fgets(s.name, sizeof(s.name), stdin);

    printf("Enter roll number: ");
    scanf("%d", &s.roll);
    printf("Enter marks: ");
    scanf("%f", &s.marks);

    printf("Displaying Information:\n");
    printf("Name: ");
    printf("%s", s.name);
    printf("Roll number: %d\n", s.roll);
    printf("Marks: %.1f\n", s.marks);

    return 0;
}

输出：

Enter information:
Enter name: Jack
Enter roll number: 23
Enter marks: 34.5
Displaying Information:
Name: Jack
Roll number: 23
Marks: 34.5

在此程序中，创建了一个学生结构体。该结构有三个成员：name (string)，roll (integer) 和 marks (float)。

然后，创建一个结构变量s来存储信息并将其显示在屏幕上。

Add two distances (in inch-feet)

12英寸等于1英尺。

#include 

struct Distance {
     
   int feet;
   float inch;
} d1, d2, result;

int main() {
     
   // take first distance input
   printf("Enter 1st distance\n");
   printf("Enter feet: ");
   scanf("%d", &d1.feet);
   printf("Enter inch: ");
   scanf("%f", &d1.inch);
 
   // take second distance input
   printf("\nEnter 2nd distance\n");
   printf("Enter feet: ");
   scanf("%d", &d2.feet);
   printf("Enter inch: ");
   scanf("%f", &d2.inch);
   
   // adding distances
   result.feet = d1.feet + d2.feet;
   result.inch = d1.inch + d2.inch;

   // convert inches to feet if greater than 12
   while (result.inch >= 12.0) {
     
      result.inch = result.inch - 12.0;
      ++result.feet;
   }
   printf("\nSum of distances = %d\'-%.1f\"", result.feet, result.inch);
   return 0;
}

输出：

Enter 1st distance
Enter feet: 23
Enter inch: 8.6

Enter 2nd distance
Enter feet: 34
Enter inch: 2.4

Sum of distances = 57'-11.0"

在此程序中，定义了一个结构距离。 该结构具有两个成员：

feet - integer
inch - float

创建了结构体Distance的两个变量d1和d2。 这些变量以英尺和英寸为单位存储距离。

然后，计算这两个距离的总和并将其存储在结果变量中。最后，把结果打印在屏幕上。

Add two complex numbers by passing structures to a function

#include 
typedef struct complex {
     
    float real;
    float imag;
} complex;

complex add(complex n1, complex n2);

int main() {
     
    complex n1, n2, result;

    printf("For 1st complex number \n");
    printf("Enter the real and imaginary parts: ");
    scanf("%f %f", &n1.real, &n1.imag);
    printf("\nFor 2nd complex number \n");
    printf("Enter the real and imaginary parts: ");
    scanf("%f %f", &n2.real, &n2.imag);

    result = add(n1, n2);

    printf("Sum = %.1f + %.1fi", result.real, result.imag);
    return 0;
}

complex add(complex n1, complex n2) {
     
    complex temp;
    temp.real = n1.real + n2.real;
    temp.imag = n1.imag + n2.imag;
    return (temp);
}

输出：

For 1st complex number
Enter the real and imaginary parts: 2.1
-2.3

For 2nd complex number
Enter the real and imaginary parts: 5.6
23.2
Sum = 7.7 + 20.9i

在此程序中，声明了一个名为complex的结构体。 它有两个成员：real和imag。 然后，我们从该结构中创建了两个变量n1和n2。

这两个结构体变量被传递给add（）函数。 该函数计算总和并返回包含该总和的结构体。

最后，在main（）函数中打印出复数的总和。

Calculate the difference between two time periods

#include 
struct TIME {
     
   int seconds;
   int minutes;
   int hours;
};

void differenceBetweenTimePeriod(struct TIME t1,
                                 struct TIME t2,
                                 struct TIME *diff);

int main() {
     
   struct TIME startTime, stopTime, diff;

   printf("Enter the start time. \n");
   printf("Enter hours, minutes and seconds: ");
   scanf("%d %d %d", &startTime.hours,
         &startTime.minutes,
         &startTime.seconds);

   printf("Enter the stop time. \n");
   printf("Enter hours, minutes and seconds: ");
   scanf("%d %d %d", &stopTime.hours,
         &stopTime.minutes,
         &stopTime.seconds);

   // Difference between start and stop time
   differenceBetweenTimePeriod(startTime, stopTime, &diff);
   printf("\nTime Difference: %d:%d:%d - ", startTime.hours,
          startTime.minutes,
          startTime.seconds);
   printf("%d:%d:%d ", stopTime.hours,
          stopTime.minutes,
          stopTime.seconds);
   printf("= %d:%d:%d\n", diff.hours,
          diff.minutes,
          diff.seconds);
   return 0;
}

// Computes difference between time periods
void differenceBetweenTimePeriod(struct TIME start,
                                 struct TIME stop,
                                 struct TIME *diff) {
     
   while (stop.seconds > start.seconds) {
     
      --start.minutes;
      start.seconds += 60;
   }
   diff->seconds = start.seconds - stop.seconds;
   while (stop.minutes > start.minutes) {
     
      --start.hours;
      start.minutes += 60;
   }
   diff->minutes = start.minutes - stop.minutes;
   diff->hours = start.hours - stop.hours;
}

输出：

Enter the start time.
Enter hours, minutes and seconds: 13
34
55
Enter the stop time.
Enter hours, minutes and seconds: 8
12
15

Time Difference: 13:34:55 - 8:12:15 = 5:22:40

在此程序中，要求用户输入两个时间段，这两个时间段分别存储在结构体变量startTime和stopTime中。

然后，函数differenceBetweenTimePeriod（）计算时间段之间的差。 在main（）函数中显示结果而不返回它（使用引用技术调用）。

Store information of 10 students using structures

#include 
struct student {
     
    char firstName[50];
    int roll;
    float marks;
} s[10];

int main() {
     
    int i;
    printf("Enter information of students:\n");

    // storing information
    for (i = 0; i < 5; ++i) {
     
        s[i].roll = i + 1;
        printf("\nFor roll number%d,\n", s[i].roll);
        printf("Enter first name: ");
        scanf("%s", s[i].firstName);
        printf("Enter marks: ");
        scanf("%f", &s[i].marks);
    }
    printf("Displaying Information:\n\n");

    // displaying information
    for (i = 0; i < 5; ++i) {
     
        printf("\nRoll number: %d\n", i + 1);
        printf("First name: ");
        puts(s[i].firstName);
        printf("Marks: %.1f", s[i].marks);
        printf("\n");
    }
    return 0;
}

输出：

Enter information of students: 

For roll number1,
Enter name: Tom
Enter marks: 98

For roll number2,
Enter name: Jerry
Enter marks: 89
.
.
.
Displaying Information:

Roll number: 1
Name: Tom
Marks: 98
.
.
.

在此程序中，将创建一个学生结构体。 该结构具有三个成员：name (string), roll (integer) and marks (float)。

然后，我们创建了具有5个元素的结构体数组s，以存储5个学生的信息。

该程序使用for循环，从用户处获取5名学生的信息，并将其存储在结构体数组中。 然后使用另一个for循环，将用户输入的信息将显示在屏幕上。

Store information of n students using structures

该程序要求用户存储noOfRecords的值，并使用malloc（）函数动态地为noOfRecords结构体变量分配内存。

#include 
#include 
struct course {
     
  int marks;
  char subject[30];
};

int main() {
     
  struct course *ptr;
  int noOfRecords;
  printf("Enter the number of records: ");
  scanf("%d", &noOfRecords);

  // Memory allocation for noOfRecords structures
  ptr = (struct course *)malloc(noOfRecords * sizeof(struct course));
  for (int i = 0; i < noOfRecords; ++i) {
     
    printf("Enter subject and marks:\n");
    scanf("%s %d", (ptr + i)->subject, &(ptr + i)->marks);
  }

  printf("Displaying Information:\n");
  for (int i = 0; i < noOfRecords; ++i) {
     
    printf("%s\t%d\n", (ptr + i)->subject, (ptr + i)->marks);
  }

  free(ptr);

  return 0;
}

输出：

Enter the number of records: 2
Enter subject and marks:
Science 82
Enter subject and marks:
DSA 73

Displaying Information:
Science     82
DSA     73

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