使用指针调用函数不会更改该指针的值？

Question

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 return 到 main 并调用函数 "ReportAnimation" 时，我无法打印该值并且我的程序崩溃了？为什么当我分配那个对象字段时它没有全局更改而只在本地函数中更改？

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);

}

我想要 initAnimation 函数更改 Animation 结构的字段我希望 reportAnimation 函数打印出该字段，证明它已更改

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.
}

如果你想在调用者中改变一个变量，你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;
}

使用指针调用函数不会更改该指针的值？

Calling a function with a pointer does not change value of that pointer?

c

malloc

struct

pointers

function-pointers