Question

我无法弄清楚如何动态地将内存分配给数组结构。我需要使用我定义的结构数组，以便我可以迭代它来打印信息。但是，我也被要求动态地为数组分配内存。

问题是当我在代码中使用malloc（）时（你会看到它被注释掉）它会从数组索引中抛出指针。有没有办法在指向数组指示存储数据的同时动态分配内存？

免责声明：我是C的新手，如果我的编码伤到你的眼睛，我会提前道歉。只是想了解程序而不是快速完成它。

计划说明：

编写一个使用＆＃39; struct＆＃39;的程序。定义包含以下学生的结构：

信息：

首字母[字符]
年龄[整数]
Id [整数]
成绩[字符]

您的课程应为5名学生打印上述信息。

提示：使用结构数组并遍历数组以打印信息。

使用以下内容创建结构

int
一个名为tag的字符指针。该指针应指向字符串的地址。

检查内存是否可用，然后为您的结构动态分配所需的内存。

为结构变量赋值并将其打印到控制台。

代码：

# include <stdio.h>
# include <string.h>
# include <stdlib.h>

int main (void)
{
    struct student  
    {
        char initial [21];
        int age;
        float id;
        char grade [3];
    } list[5];

    struct student * tag = list;

    //LINE BELOW IS WHAT I WAS TRYING TO DO; BUT THROWS POINTER OFF OF STRUCT ARRAY
    //tag = ( struct student * ) malloc ( sizeof ( struct student ));
    strcpy(tag->initial, "KJ");
    tag->age = 21;
    tag->id = 1.0;
    strcpy (tag->grade, "A");
    tag++;

    strcpy(tag->initial, "MJ");
    tag->age = 55;
    tag->id = 1.1;
    strcpy (tag->grade, "B");
    tag++;

    strcpy(tag->initial, "CJ");
    tag->age = 67;
    tag->id = 1.2;
    strcpy (tag->grade, "C");
    tag++;

    strcpy(tag->initial, "SJ");
    tag->age = 24;
    tag->id = 1.3;
    strcpy (tag->grade, "D");
    tag++;

    strcpy(tag->initial, "DJ");
    tag->age = 27;
    tag->id = 1.4;
    strcpy (tag->grade, "F");
    tag++;

    int n;

    for ( n = 0; n < 5; n++ ) {
            printf ( "%s is %d, id is %f, grade is %s\n", 
              list [ n ].initial, list [ n ].age, list [ n ].id, list [ n ].grade);
            }

    return 0;

}

Answer 1

试试这个....这里我宣布一个结构指针tag。我为5个学生结构分配内存。分配值后，现在tag指向最后一个值。所以我将它递减5次。这个程序只使用一个指针。如果你想，你可以尝试使用指针数组。

我在程序中提到了//changes

的更改

# include <stdio.h>
# include <string.h>
# include <stdlib.h>

int main (void)
{
 struct student
  {
    char initial [21];
    int age;
    float id;
    char grade [3];
  } list[5];

struct student * tag;


    tag = ( struct student * ) malloc (5* sizeof (struct student));//changes

strcpy(tag->initial, "KJ");
tag->age = 21;
tag->id = 1.0;
strcpy (tag->grade, "A");
tag++;

strcpy(tag->initial, "MJ");
tag->age = 55;
tag->id = 1.1;
strcpy (tag->grade, "B");
tag++;

strcpy(tag->initial, "CJ");
tag->age = 67;
tag->id = 1.2;
strcpy (tag->grade, "C");
tag++;

strcpy(tag->initial, "SJ");
tag->age = 24;
tag->id = 1.3;
strcpy (tag->grade, "D");
tag++;

strcpy(tag->initial, "DJ");
tag->age = 27;
tag->id = 1.4;
strcpy (tag->grade, "F");
tag++;

int n;
    tag=tag-5;//changes

for ( n = 0; n < 5; n++ ) {
        printf ( "%s is %d, id is %f, grade is %s\n",
          tag->initial, tag->age, tag->id, tag->grade);
            tag++;//changes
        }

return 0;
}

使用pinter数组...（而不是使用单独的赋值，你可以使用循环为每个学生分配值）

    #include <stdio.h>
    #include <string.h>
    #include <stdlib.h>

    int main (void)
    {
            struct student
            {
             char initial [21];
             int age;
             float id;
             char grade [3];
            } list[5];

        struct student *tag[5];
    int i;

    for(i=0;i<5;i++)
    tag[i]= ( struct student * ) malloc (sizeof (struct student));

        strcpy(tag[0]->initial, "KJ");
        tag[0]->age = 21;
        tag[0]->id = 1.0;
        strcpy (tag[0]->grade, "A");

        strcpy(tag[1]->initial, "MJ");
        tag[1]->age = 55;
        tag[1]->id = 1.1;
        strcpy (tag[1]->grade, "B");

        strcpy(tag[2]->initial, "CJ");
        tag[2]->age = 67;
        tag[2]->id = 1.2;
        strcpy (tag[2]->grade, "C");

        strcpy(tag[3]->initial, "SJ");
        tag[3]->age = 24;
        tag[3]->id = 1.3;
        strcpy (tag[3]->grade, "D");

        strcpy(tag[4]->initial, "DJ");
        tag[4]->age = 27;
        tag[4]->id = 1.4;
        strcpy (tag[4]->grade, "F");

        for ( i = 0; i < 5; i++ )
        {
                printf ( "%s is %d, id is %f, grade is %s\n",
             tag[i]->initial, tag[i]->age, tag[i]->id, tag[i]->grade);

        }

        return 0;

     }

Answer 2

如果您使用指针来读取和打印结构值，则不需要数组。如果您尝试使用数组，则可以使用此方法。

 #include <stdio.h>
    # include <string.h>
    # include <stdlib.h>

    int main (void)
    {
       struct student  
       {   
          char initial [21];
          int age;
          float id; 
          char grade [3];
       } list[5];

       struct student *tag = ( struct student * ) malloc ( sizeof ( struct student ) * 5);

       int i;
       for(i=0;i<5;i++)
       {   
          printf("Enter the student initial\n");
          scanf("%s",list[i].initial);
          printf("Enter the student age\n");
          scanf("%d",&list[i].age);
          printf("Enter the student id\n");
          scanf("%f",&list[i].id);
          printf("Enter the grade\n");
          scanf("%s",list[i].grade);
          tag++;
       }   

       int n;

       for ( n = 0; n < 5; n++ ) { 
          printf ( "%s is %d, id is %f, grade is %s \n", 
                list [ n ].initial, list [ n ].age, list [ n ].id, list [ n ].grade);
       }   

       return 0;

    }

结构数组中的动态分配内存（C中）

2 个答案: