skp6.c: In function ‘initList’: skp6.c:48: error: expected expression before ‘)’ toke

[opcs]

here is my complete code, can any one help me why does this error occurs and how to resolve it.

also, error is at the line "list.hdr =(list *) malloc(sizeof(struct SkipList));" is the size allocation here is correct or it should be diffrent.

code:

#include <stdio.h>
#include <stdlib.h>
#define compLT(a,b) (a < b)
#define compEQ(a,b) (a == b)
#define MAXLEVEL 15
// levels range from (0 .. MAXLEVEL)
struct Node_ {
    int data;
    struct Node_ *forward[1];
} Node;
struct SkipList{
 struct Node_ *hdr;                  // list Header
  int listLevel;              // current level of list
}list;
 
<snipped>
 
//  initialize skip list
   int initList() 
  {
    int i; 
    list.hdr =(list *) malloc(sizeof(struct SkipList));
    for (i = 0; i <= MAXLEVEL; i++)
    list.hdr->forward[i] = list.hdr;
    list.listLevel = 0;
    printf("\nskip initialized");
    printf("\ninitial current level of list=%d",list.listLevel);
    printf("\ninitial skip is empty and all hdr pointer=list.hdr");
  }
 

thank you !

[Banfa]

list is a variable of type struct SkipList.

On the line in question you try to use it to cast the output of malloc.

In fact you do this

Reserve memory sized to the struct SkipList
Attempt to cast it to the variable "list *" this is a complete syntax error
Allocate the result to a pointer of type Node_ *

This is a completely confused attempt at allocation.

For some type T then you should perform a simple allocation like this (in C)

T *ptrData;
ptrData = malloc(sizeof *ptrData);
 

Some people use the alternative

T *ptrData;
ptrData = malloc(sizeof(T));
 

I prefer the first form because if for some reason you change the type of ptrData you do not have to change all your allocations.

In C++ you have to cast the output of malloc so it becomes

T *ptrData;
ptrData = (T *)malloc(sizeof(T));
 

but then you should be using new in C++.