来自不兼容指针类型的赋值[Wincompatible-pointer-types]

Question

For a testing assignment I am supposed to detect errors in the following C program:

    #include <stdlib.h>
    #include <stdio.h>
    typedef struct {
    struct chain *next;       
    int contents;  } chain; 

    int main()  {     
        int index;     
        chain *list;     
        chain *p;    
        chain *pointer;     
        list = malloc(sizeof(chain));     
        p = list;     

        for(index=0;index<10;index++) {       
            (*p).contents = index;       
            (*p).next = malloc(sizeof(chain));       
            p = (*p).next; 
           } ;     
        p = pointer = list;     
        index = 0;     

        while (index < 9) { 
            printf("cell # %d: %d\n",index,(*p).contents);       
            p = (*p).next;       
            free(pointer);       
            pointer = p;       
            index++; 
          } ;     
        printf("First cell: %d\n",(*list).contents);     
        return 0;  
      }

I get the following 2 errors:

Assignment from incompatible pointer type [Wincompatible-pointer-types]

at the statements p=(*p).next .

I have a feeling this might be something trivial, but I am a complete novice at C and can't figure out the correct syntax for these statements. Help much appreciated.

Answer 1

typedef struct {
    struct chain *next;
    int contents;
} chain;

this creates an anonymous structure chain . But the next pointer is of type struct chain* . But there's no type struct chain defined in your code at all! So the compiler can't know the relationship between the type chain and struct chain .

You can instead name the struct:

typedef struct chain {
    struct chain *next;
    int contents;
} chain;