用指针调用函数不会改变该指针的值？

Question

这个问题已经有答案了:

Pointers as function arguments in C (7 个回答)

已关闭 3 年前。

当我调用“InitAnimation”函数时，我传递了对象 RG 的地址。当我分配该对象的“animationName”字段时，我可以成功打印该字段。但是当我返回主函数并调用函数“ReportAnimation”时，我无法打印该值并且我的程序崩溃了？为什么当我分配该对象字段时，它不会全局更改，而仅在本地函数中更改？

我也尝试过为animationName 字段分配内存，但这不起作用。

struct Frame {
    char* frameName;
    struct Frame* pNext;
};

typedef struct {
    char* animationName;
    struct Frame* frames;
}Animation;


int main(void) {

    char response;

    BOOL RUNNING = TRUE;

    Animation RG;

    InitAnimation(&RG);

    while (RUNNING) {
        printf("MENU\n Enter 1 to ReportAnimation\n");
        scanf("%c", &response);

        switch (response) {
        case '1':InsertFrame(&RG); 
        break;
    }
    }

    return 0;
}

void InitAnimation(Animation* pointer) {

    pointer = (Animation*)malloc(sizeof(Animation));

    char* input;
    input = (char*)malloc(sizeof(input));

    printf("Please enter the Animation name:");
    fgets(input, 32, stdin);

    //pointer->animationName = (char*)malloc(sizeof(char)*10);

    //Setting animation name
    pointer->animationName = input;


    //This print function works
    printf("\nThe name is %s", pointer->animationName);

}

void ReportAnimation(Animation* pointer) {

    //This print function does not work
    printf("Animation name is %s\n", pointer->animationName);

}

我希望 initAnimation 函数更改 Animation 结构的字段我希望 reportAnimation 函数打印出该字段，证明其已更改

Answer 1

更改函数变量的值(包括声明为参数的变量)对调用者没有影响。

void bad1(int i) {
   i = 1;  // No effect on the caller.
}

void bad2(void* p) {
   p = NULL;  // No effect on the caller.
}

如果您想更改调用者中的变量，则需要传递一个指向它的指针。

void good1(int* i_ptr) {
   *i_ptr = 1;
}

void good2(void** p_ptr) {
   *p_ptr = NULL;
}

因此，要么将指针传递给指针，要么将指针传递给已分配的结构。你可以使用

Animation ani;
Animation_init(&ani, name);
...
Frame* frame = Frame_new(name);
Animation_append_frame(&ani, frame);
...
Animation_destroy(&ani);

或

Animation* ani = Animation_new(name);
...
Frame* frame = Frame_new(name);
Animation_append_frame(ani, frame);
...
Animation_delete(ani);

假设你有

typedef struct Frame {
    char* name;
    struct Frame* next;
} Frame;

typedef struct {
    char* name;
    Frame* first_frame;
    Frame* last_frame;
} Animation;

void Animation_init(Animation* self, const char* name) {
    self->name = strdup(name);
    self->first_frame = NULL;
    self->last_frame = NULL;
}

void Animation_destroy(Animation* self) {
    Frame* head = self->first_frame;
    while (head != NULL) {
       Frame* next = head->next;
       Frame_delete(head);
       head = next;
    }

    free(self->name);
}

Animation* Animation_new(const char* name) {
    Animation* self = malloc(sizeof(Animation));
    Animation_init(self, name);
    return self;
}

void Animation_delete(Animation* self) {
    Animation_destroy(self);
    free(self);
}

void Animation_append_frame(Animation* self, Frame* frame) {
   if (self->last_frame == NULL) {
      self->first_frame = frame;
   } else {
      self->last_frame->next = frame;
   }

   while (frame->next != NULL)
      frame = frame->next;

   self->last_frame = frame;
}