C编程：‘sizeof’对不完整类型‘Tipo {aka struct ElemSCl}的应用无效

Question

我正在学习，这是一个简单的测试。

scl.h

#ifndef sclh
#define sclh

typedef struct{
    int value; }listnote;


 struct ElemSCL {
       listnote info;
      struct ElemSCL*next;
                 };
                 
typedef struct ElemSCl Tipo;
typedef Tipo *Mangiato;


void Addscl(Mangiato*scl, int e) ;
#endif

prove.c

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

//create node
void Addscl (Mangiato *scl, int e) {
 Mangiato temp;
 temp = *scl;
*scl= (Tipo*) malloc(sizeof(Tipo)); 
*(scl)->info.value= e;
*(scl)->next = temp;
   }

Main.c

#include<stdio.h>
#include<stdlib.h>
#include"scl.h"
int main()
{
    Tipo *scl= NULL;
    Addscl (&scl,3);
  printf("%d",  *(scl)->info.value);
      
        
    }
 return 0;
}

我得到以下错误：

main.c:9:22: error: dereferencing pointer to incomplete type ‘Tipo {aka struct ElemSCl}’
 printf(" %d",  *(scl)->info.value);
                      ^~
prove.c: In function ‘Addscl’:
prove.c:9:29: error: invalid application of ‘sizeof’ to incomplete type ‘Tipo {aka struct ElemSCl}’
 *scl= (Tipo*) malloc(sizeof(Tipo));
                             ^~~~
prove.c:10:7: error: ‘*scl’ is a pointer; did you mean to use ‘->’?
 *(scl)->info.value= e;
       ^~
       ->
prove.c:11:7: error: ‘*scl’ is a pointer; did you mean to use ‘->’?
 *(scl)->next = temp;
       ^~
       ->

Answer 1

表达式*(scl)->info.value与*(scl->info.value)相同。也就是说，取消对value成员的引用。

您需要(*scl)->info.value来解除对scl指针的引用。

至于sizeof问题，您有

 struct ElemSCL { ... };

和

typedef struct ElemSCl Tipo;

密切注意拼写...在typedef中使用小写的l，应该是大写的L

typedef struct ElemSCL Tipo;