“名称”的存储大小;未知
编译此 .c 源文件时出现此错误
/INIT_SOURCE_BUILD/src/names_list.c:7: 错误:“名称”的存储大小不是 已知
#include <stdio.h>
#include "list.h"
int main(){
struct List names;
names->size = 3;
struct ListElmt michael;
struct ListElmt john;
struct ListElmt adams;
names->head = michael;
michael->data = 12;
michael->next = john;
john->data = 14;
john->next = adams;
adams->data = 16;
struct ListElmt pointer = List->head;
for(int x = 0; x < 3 ; x++){
printf("Iteration.%d data: %d", x, pointer->data);
pointer->next = pointer->next->next;
}
}
,这是该链表的标头
#ifndef LIST_H
#define LIST_H
#include <stdio.h>
/* Define linked list elements*/
typedef struct _ListElmt{
void *data;
struct _ListElmt *next;
} ListElmt;
/* Define a structure for the list*/
typedef struct _List{
int size;
int (*match)(const void *key1, const void *key2);
void (*destroy)(void *data);
ListElmt *head;
ListElmt *tail;
} List;
void list_init(List *list, void (*destroy)(void *data));
void list_destroy(List *list);
int list_ins_next(List *list, ListElmt *element, const void *data);
int list_rem_next(List *list, ListElmt *element, void **data);
int list_size(const List *list);
ListElmt *list_head(const List *list);
ListElmt *list_tail(const List *list);
int list_is_head(const ListElmt *element);
int list_is_tail(const ListElmt *element);
void *list_data(const ListElmt *element);
ListElmt *list_next(const ListElmt *element);
#endif
I get this error while compiling this .c source file
/INIT_SOURCE_BUILD/src/names_list.c:7:
error: storage size of ‘names’ isn’t
known
#include <stdio.h>
#include "list.h"
int main(){
struct List names;
names->size = 3;
struct ListElmt michael;
struct ListElmt john;
struct ListElmt adams;
names->head = michael;
michael->data = 12;
michael->next = john;
john->data = 14;
john->next = adams;
adams->data = 16;
struct ListElmt pointer = List->head;
for(int x = 0; x < 3 ; x++){
printf("Iteration.%d data: %d", x, pointer->data);
pointer->next = pointer->next->next;
}
}
and here is header of this linked list
#ifndef LIST_H
#define LIST_H
#include <stdio.h>
/* Define linked list elements*/
typedef struct _ListElmt{
void *data;
struct _ListElmt *next;
} ListElmt;
/* Define a structure for the list*/
typedef struct _List{
int size;
int (*match)(const void *key1, const void *key2);
void (*destroy)(void *data);
ListElmt *head;
ListElmt *tail;
} List;
void list_init(List *list, void (*destroy)(void *data));
void list_destroy(List *list);
int list_ins_next(List *list, ListElmt *element, const void *data);
int list_rem_next(List *list, ListElmt *element, void **data);
int list_size(const List *list);
ListElmt *list_head(const List *list);
ListElmt *list_tail(const List *list);
int list_is_head(const ListElmt *element);
int list_is_tail(const ListElmt *element);
void *list_data(const ListElmt *element);
ListElmt *list_next(const ListElmt *element);
#endif
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评论(3)
当你像这样
typedef
一个struct
时,你在声明它时不必使用struct
:而不是
它也不是一个指针,所以
names->size
应该是names.size
。When you
typedef
astruct
like that, you don't have to usestruct
when declaring it:instead of
It's also not a pointer, so
names->size
should benames.size
.struct List name;
没有声明指针,但您试图取消引用它(使用->
)。请改用names.size
。struct List names;
doesn't declare a pointer, but you're trying to dereference it (using->
). Usenames.size
instead.该结构称为_List。 typedef 是 List。所以你想要
或者
你的行正在声明一个尚未定义的“结构列表”。
关于 的其他答案非常正确。对比->问题
The struct is called _List. The typedef is List. So you want
or
Your line is declaring a "struct List" which isn't defined yet.
The other answers are quite correct about the . vs -> issue