C语言 “sizeof”对不完整类型列表结构 C 的无效应用

invalid application of 'sizeof' to incomplete type list struct C

提问by joseph

I'm trying to implement an replacement algorithm that deals with page faults. So i'm trying to creat a circular linked list using malloc and im getting the following error: "invalid application of sizeof' to incomplete typepageInMemory'.following is the code:

我正在尝试实现一种处理页面错误的替换算法。因此，我正在尝试使用 malloc 创建一个循环链表，但出现以下错误：“ sizeof' to incomplete typepageInMemory 的应用程序无效。以下是代码：

 typedef  struct {

            int use; 
            int reference;
            int free;
            struct pageInMemory* next;
            } pageInMemory;


            int main()
            {

                int i;
                struct pageInMemory* start, *nn, *temp, *hand;
                start = NULL;

                    for(i=0; i< pNum; i++) 
                    {
                nn = (struct pageInMemory *)malloc(sizeof(struct pageInMemory));
                        nn->use = 0;
                        nn->free = 1;

                        if(start==NULL)
                        {
                            nn->next = nn;
                            start =nn;
                        }

                        else
                        {     // sfhsdifhsdifj sdijfjsd 
                            temp = start;
                            while(temp->next != start)
                            {
                                temp = temp->next; 
                            }

                            temp->next = nn;
                            nn->next = start;
                            start = nn;

                        }   

                    }



                hand = start;
                temp = start;

             while(temp->next != start->next)
             {
                printf("%d\n", temp->use); //hi
             } 


                return 0;// bye
             }  

so am i not supposed to use malloc this way ?

所以我不应该这样使用 malloc 吗？

回答by Dayal rai

change your struct definition as

将您的结构定义更改为

struct pageInMemory{

            int use;
            int reference;
            int free;
            struct pageInMemory* next;
            };

to get your code working. And just for your info do not typecast void* coming from malloc.

让你的代码工作。并且只是为了您的信息，不要对来自 malloc 的 void* 进行类型转换。

回答by alk

It should be either:

它应该是：

typedef  struct pageInMemory_s {

        int use; 
        int reference;
        int free;
        struct pageInMemory_s * next;
        } pageInMemory;

pageInMemory* start, *nn, *temp, *hand;
... 
nn = malloc(sizeof(pageInMemory));

or

或者

struct pageInMemory {

        int use; 
        int reference;
        int free;
        struct pageInMemory* next;
        };

struct pageInMemory* start, *nn, *temp, *hand;
... 
nn = malloc(sizeof(struct pageInMemory));

A third variant would be:

第三种变体是：

typedef  struct pageInMemory {

        int use; 
        int reference;
        int free;
        struct pageInMemory * next;
        } pageInMemory;

pageInMemory* start, *nn, *temp, *hand;
... 

For this 3rd option you can use:

对于第三个选项，您可以使用：

nn = malloc(sizeof(pageInMemory));

or

或者

nn = malloc(sizeof(struct pageInMemory));

This latter variation I feel is very irritating as there is one name for two differentthings:

我觉得后一种变化非常令人恼火，因为两种不同的事物有一个名称：

• The structure definition/declaration struct pageInMemory
• The type definition/declaration pageInMemory

• 结构定义/声明 struct pageInMemory
• 类型定义/声明 pageInMemory

I would not recommend to use this 3rd option, but the second.

我不建议使用第三个选项，而是第二个。

回答by VoidPointer

pageInMemoryis itself defined as type. So instead of this,

pageInMemory本身被定义为类型。所以，而不是这个，

nn = (struct pageInMemory *)malloc(sizeof(struct pageInMemory))

Use this,

用这个，

nn = (pageInMemory *)malloc(sizeof(pageInMemory))