结构体实例

在这里插入图片描述

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: JackEnter roll number: 23Enter marks: 34.5Displaying Information:Name: JackRoll number: 23Marks: 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 distanceEnter feet: 23Enter inch: 8.6Enter 2nd distanceEnter feet: 34Enter inch: 2.4Sum 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 numberEnter the real and imaginary parts: 2.1-2.3For 2nd complex numberEnter the real and imaginary parts: 5.623.2Sum = 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 periodsvoid 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: 133455Enter the stop time.Enter hours, minutes and seconds: 81215Time 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: TomEnter marks: 98For roll number2,Enter name: JerryEnter marks: 89...Displaying Information:Roll number: 1Name: TomMarks: 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: 2Enter subject and marks:Science 82Enter subject and marks:DSA 73Displaying Information:Science     82DSA     73

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