adding a Node using a function...why is this happening?

**TonyG** · 05-09-2011

I have run this code through the debugger and i can see the "menu->headInventory" get populated with the data from the "new" node. But as soon as it leaves the function the data stored in the "menu->headInventory" is no longer accessible, Im passing by reference so why is this occurring?

Code:

 if ((validation = addNode(&menu, &lineBuffer, &new)) != TRUE) {
            fprintf(stderr, "Unable to add to node.\n");
            exit(EXIT_FAILURE);
        }

Code:

int addNode(HeadNode *menu, char *lineBuffer, InventoryTypePtr new) {

    char *tokens = NULL, delims[] = "|", type;
    int count = 1; /*case 1*/

    tokens = strtok(lineBuffer, delims);

    new = (InventoryTypePtr)malloc(sizeof(InventoryTypePtr));
    if (new == NULL) {
        fprintf(stderr, "Memory allocation for current failed");
        return FALSE;
    }

    /* Populate the node with the imported information from the
    line buffer */
    while (tokens != NULL) {
        printf("%s", tokens);
        switch (count) {
        case 1 :
            strcpy(new->inventoryID, tokens);
            break;
        case 2 :
            type = *tokens;
            new->foodType = type;
            break;
        case 3 :
            strcpy(new->inventoryName, tokens);
            break;
        case 4 :
            strcpy(new->inventoryDescription, tokens);
            break;
        }
        count++;
        tokens = strtok( NULL, delims);
    }
    new->nextInventory = menu->headInventory;
    menu->headInventory = new;
    menu->numInventoryNodes += 1;
    return TRUE;
}

**MK27** · 05-09-2011

Originally Posted by TonyG

Code:

    new = (InventoryTypePtr)malloc(sizeof(InventoryTypePtr));

Because of it's name and the way you use it, I presume "InventoryTypePtr" was typedef'd this way:

Code:

typedef struct Inventory* InventoryTypePtr;

IMO, typedefing pointers is not a good practice because it serves no purpose and could easily lead to confusion, but that's just my opinion.

Anyway, since InventoryTypePtr is a pointer, how much memory do you think that malloc assigns? Throw this in:

Code:

printf("Size of InventoryTypePtr: %d\n", sizeof(InventoryTypePtr));

I think you will find this does not jibe well with the actual size of the struct that a "InventoryTypePtr" points to.

It seems, however, that you are getting off lucky; add a bit more to the program or compile it on another computer and this discrepancy will cause an access violation and seg fault, or -- even worse -- overwrite memory belonging to the program but not that variable, causing weirder and harder to solve problems.

The reason your pointer is not populated outside that function is because you are reassigning the value of a value passed in. Look:

Code:

void example (int *x) {
      x = malloc(whatever);
}

Is exactly the same as:

Code:

void example (int x) {
      x = whatever;
}

Do you see now what is going on? Now consider:

Code:

void example (int *x) {
      *x = whatever;
}

Of those three examples, only the last one will affect the value of x in the caller. To assign memory to a passed in pointer and make it meaningful to the caller, you need to use a pointer to a pointer:

Code:

void example (int **x) {
       *x = malloc(sizeof(int));
       **x = whatever;
}

**TonyG** · 05-09-2011

Great i understand now. Thank you.
Just one thing i would like to clear up. When using malloc. Would this be the correct way of writing it?

Code:

new = (InventoryTypePtr)malloc(sizeof(InventoryType));
 /* or better still */
new = (InventoryType *)malloc(sizeof(InventoryType));

Thanks heaps for the help.

**anduril462** · 05-09-2011

The closer one would be

Code:

new = (InventoryType *)malloc(sizeof(InventoryType));

since it avoids typedef'ing pointers, which as MK27 pointed out, is confusing/problematic. But better yet, don't cast malloc.

**TonyG** · 05-10-2011

Originally Posted by MK27

Code:

void example (int **x) {
       *x = malloc(sizeof(int));
       **x = whatever;
}

How do i use this pointer to a pointer. If it is a struct and i want to populate it as with my initial post?

Code:

strcpy(new->inventoryID, token);

**rags_to_riches** · 05-10-2011

You can make this easier to use like this:

Code:

void example (int **x) {
    int *local_x = malloc(sizeof(*local_x));
    /* Work with local_x, it's just a pointer now */
    
    /* Assign address of local_x to *x for return */
    *x = local_x;
}

**TonyG** · 05-10-2011

Great thank you

**MK27** · 05-10-2011

Originally Posted by TonyG

How do i use this pointer to a pointer. If it is a struct and i want to populate it as with my initial post?

Code:

strcpy(new->inventoryID, token);

Code:

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

struct eg {
	int x;
	char data[256];
};

int create_eg (struct eg **new) {
	*new = malloc(sizeof(struct eg));
	if (!new) return -1;
// dereferencing (*) a "pointer to a pointer" yields a normal pointer:
	(*new)->x = 666;
	strcpy((*new)->data, "hello world");  
	return 0;
}

int main(void) {
	struct eg *test;   // a pointer

// the address of (&) a pointer is a pointer to a pointer:
	create_eg(&test);    

	printf("%d %s\n", test->x, test->data)

	free(test);

	return 0;
}

**TonyG** · 05-10-2011

Thanks guys. Managed to get both approaches to work. For readability i have used the second. But both were effect and solutions to my problem.

**TonyG** · 05-10-2011

Still having trouble understanding passing a struct as a pointer "call by reference".

menu has been declared as a HeadNode struct

Code:

typedef struct head
{
   InventoryTypePtr headInventory;
   unsigned numInv;
} HeadNode;

an instance of HeadNode is declared in my Main as

Code:

 HeadNode menu;

menu is being passed through functions as a pointer

Code:

addNode(&menu, ....);

Code:

addNode(*menu, ...);

From the above examples i have the information stored in headInventory and headInventory's follow on nodes.
But if i try and set the value of numInv = 0 in a function. Once it leaves the function the value of numInv is set to <large rand num>.

**tabstop** · 05-10-2011

You can't have both "addNode(&menu, ....);" and "addNode(*menu, ...);" at the same time. The first would be correct.

**TonyG** · 05-10-2011

sorry didn't make the second very clear should have been

Code:

addNode(*menu, ...) {
menu->numInv = 0;
}

**tabstop** · 05-10-2011

Your idea of snippet is lacking. Complete example, please.

**TonyG** · 05-10-2011

in the header

Code:

typedef struct head
{
   InventoryTypePtr headInventory;
   unsigned numInv;
} HeadNode;

Code:

int main(int argc, char* argv[]) {

    HeadNode menu;
    systemInit(&menu);
    printf("menu.num = %u", menu.numInv);

Code:

void systemInit(HeadNode* menu) {
menu->numInv = 0;
}

Output is some over sized number stored in numInv...

**~~CommonTater~~** · 05-10-2011

Hi Tony... You really need to give us more than 2 and 3 line examples here... Post your code... or at least post complete functions.

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