Thread: Error in pointer conversion

I am simply trying to create a vector but as soon as i compile I get this error message

"base operand of -> has non pointer type VectorT"

I cant seem to find a way around this, any help is greatly appreciated

Code:

typedef char NodeItemT;

typedef struct
{
	NodeItemT info;
} NodeT;

typedef struct
{
	NodeT *list;
	int   capacity;
	int   size;
} VectorT;

NodeT *newNode(NodeItemT item);
VectorT *new1(int cap);

int main()
{
    
 getch();
 return 0;   
}


NodeT *newNode(NodeItemT item)
{
	NodeT* node;

	node = (NodeT*) malloc (sizeof(NodeT));
    node->info=item;
	return node;
}

VectorT *new1(int cap)
{
VectorT *init;
init=(VectorT*)malloc (15*sizeof(VectorT));

int i;
for(i=0;i<15;i++)
{
   init[i]->list->info='\0'; // Error here
   init[i]->capacity=cap; //Error here
   init[i]->size=0;     //Error here
}
   
return init;
        
}

Init is a pointer to a vector... if you want to create an array of pointers you need to use...

Code:

init=(VectorT*)malloc (15*sizeof(*init);)

// OR


init=(VectorT*)malloc (15*sizeof(*VectorT));

You want the size of the pointer... not the size of the struct.

I tried those and I still get the same error

Originally Posted by CommonTater

Code:

 init=(VectorT*)malloc (15*sizeof(*VectorT));

Methinks this wont work. One can not dereference a type name. I think you just meant sizeof(VectorT) without the *

Originally Posted by camel-man

"base operand of -> has non pointer type VectorT"

The -> is the member access operand for a pointer to a struct. So what it's saying is that whatever is on the left of a -> is not a pointer to a struct. You declare init as a VectorT *, but you treat it as an array, so init[i] gives you back a VectorT, not a pointer to a VectorT. Thus, you need

Code:

    init[i].list->info = NULL; // You don't want the nul character '\0', you probably want NULL, the null pointer constant
    init[i].capacity = cap;
    init[i].size = 0;

If i could kiss you i would

Originally Posted by camel-man

If i could kiss you i would

Flattered, but given your user name, I think I'll pass. Now, if your user name were really_hot_super_model_chick_with_no_STDs, then I might just take you up on that offer

haha ok well the offer stands