使用结构作为函数参数时出错？

Question

I made a function that has a pointer and an integer as parameters. Its supposed to print values from a linked list where the pointer points at the first object. The function looks like:

void printlist(talstrul *lank, int langd)
    {   int j;
        talstrul *temppek = lank;
            for(j=0; j<langd; j++)
            {   

        printf("%d\n",*temppek);
        temppek = temppek->next;    

        }
    }

The error I get is:

syntax error : missing ')' before '*'
syntax error : missing '{' before '*'

The struct is defined as follows:

struct talstrul
{
    int num;
    struct talstrul *next;


};
typedef struct talstrul talstrul;

Answer 1

You appear not to have defined talstrul (or included the definition here). Perhaps it's a struct (but not a typedef struct ) and you wanted struct talstrul * lank and struct talstrul * temppek = lank; .

Also this line:

printf("%d\n",*temppek);

has got to be wrong if temppek points to a struct as implied by

temppek = temppek->next;

Answer 2

您的错误几乎可以肯定地不在该函数中-您可能在页面上方有一个未终止的块。

Answer 3

You want to print the values in the linked list, ie the num entries of the structs.

printf("%d\n",temppek->num);

To catch these kinds of errors, compiling with gcc -Wall is good practice.

---

Edit: Indeed, this snippet works fine, and the problem is elsewhere.

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

typedef struct talstrul {
  int num;
  struct talstrul *next;
} talstrul;

void printlist(talstrul *lank, int langd) {
  int j;
  talstrul *temppek = lank;
  for(j=0; j<langd; j++) {
      printf("%d\n", temppek->num);
      printf("Illogical printing: %d\n",*temppek);
      temppek = temppek->next;
    }
}

int main (void) {
  talstrul *mylank = (talstrul*) malloc (sizeof (talstrul));
  mylank->num = 13;
  mylank->next = NULL;
  printlist(mylank, 1);

  free(mylank);
  return 0;
}