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SuperCarrot
Asked: 2025-03-04 23:45:03 +0800 CST

如何改变指针数组中的每个结构？

  • 772

假设我有一个指向结构的指针数组。如何使数组的每个元素都指向不同的内存位置，以便我可以通过指针单独更改这些结构值？

一些示例代码：

struct Person {
    int age;
    char *name[64];
}

int main() {
    struct Person *people[4];
    struct Person tmp_person;
    people[0] = &tmp_person;
    people[0]->age = 12;
    people[0]->name = "John";

    people[1] = &tmp_person;
    people[1]->age = 34;
    people[1]->name = "Mike";
}

这里，每个 person 元素都将相同，因为所有数组元素都指向同一内存（tmp_person）。我知道代码不正确，我只是想举个例子。那么我该如何创建唯一指针呢？为每个元素（tmp_person1、tmp_person2、...）创建一个新变量似乎很愚蠢，而且非常困难。

arrays

2 个回答

  1. Best Answer

    正如 dbush 所说：更典型的情况是，可能你（不知道你）想要的是一个对象数组，而不仅仅是指向对象的指针：

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

struct Person {
    int age;
    const char *name;
};

void print_persons(const struct Person *persons, int arrSz)
{
  for( int i=0; i<arrSz; i++)
  {
    printf("Person %s, age %d\n", persons[i].name, persons[i].age);
  }
}


int main() {
    struct Person people[2];
    people[0].age = 12;
    people[0].name = "John";

    people[1].age = 34;
    people[1].name = "Mike";

    struct Person others[]
      = { {23, "Jim"},
          {24, "Joe"},
          {25, "Jack"}
        };

    // Use pointers:
    struct Person *peoplePtrs[2];
    peoplePtrs[0] = malloc(sizeof(struct Person));
    peoplePtrs[0]->age = 12;
    peoplePtrs[0]->name = "JohnPtr";

    peoplePtrs[1] = malloc(sizeof(struct Person));
    peoplePtrs[1]->age = 34;
    peoplePtrs[1]->name = "MikePtr";

    print_persons(people, sizeof people/sizeof *people);
    print_persons(others, sizeof others/sizeof *others);
    for( int i=0; i<sizeof peoplePtrs/sizeof *peoplePtrs; i++)
    {
      printf("Person %s, age %d\n", peoplePtrs[i]->name, peoplePtrs[i]->age);
    }

    // good form to free the allocated memory, even though it is
    // not really necessary at the end of a program
    free(peoplePtrs[0]);
    free(peoplePtrs[1]);
}

    我将 的类型更正name为简单指针；它最适合静态字符串，如您的示例所示。您可以使用字符数组，但必须使用strcpy或初始化它们memcpy— 无法分配数组。

    我还向您展示了如何直接初始化数组，如果您在编程时知道数据，该方法就可以起作用，如您的示例所示。

    您当然可以使用指向结构的指针数组，如最后一部分所示；通常，指向的对象将使用 malloc 动态分配（然后释放）。

    我编写了一个小函数，只用于打印一个人以显示内容。

    • 3

  2. malloc可用于根据需要分配内存。可以
    将struct Person *people多个结构分配给指针。此示例为三个结构分配内存。

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

struct Person {
    int age;
    char name[64];
};

int main ( void) {
    struct Person *people = NULL;

    if ( NULL == ( people = malloc ( sizeof *people * 3))) {
        fprintf ( stderr, "problem malloc\n");
        return 1;
    }
    people[0].age = 12;
    strcpy ( people[0].name, "John");
    people[1].age = 34;
    strcpy ( people[1].name, "Mike");
    people[2].age = 22;
    strcpy ( people[2].name, "Alice");

    for ( int each = 0; each < 3; ++each) {
        printf ( "age %d\n", people[each].age);
        printf ( "name %s\n", people[each].name);
    }
    free ( people);
}

    使用指针数组，可以为每个指针分配内存。

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

struct Person {
    int age;
    char name[64];
};

int main ( void) {
    struct Person *people[4] = { NULL};

    for ( int each = 0; each < 4; ++each) {
        if ( NULL == ( people[each] = malloc ( sizeof (struct Person)))) {
            fprintf ( stderr, "problem malloc\n");
            while ( each) {
                --each;
                free ( people[each]);
            }
            return 1;
        }
    }
    people[0]->age = 12;
    strcpy ( people[0]->name, "John");
    people[1]->age = 34;
    strcpy ( people[1]->name, "Mike");
    people[2]->age = 22;
    strcpy ( people[2]->name, "Alice");
    people[3]->age = 28;
    strcpy ( people[3]->name, "April");

    for ( int each = 0; each < 4; ++each) {
        printf ( "age %d\n", people[each]->age);
        printf ( "name %s\n", people[each]->name);
    }
    for ( int each = 0; each < 4; ++each) {
        free ( people[each]);
    }
}
    • 0

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