CODEWITHSUNDEEP
carraysmalloc
darksky
darksky

Reputation: 21019

Dereferencing Pointer in Dynamic Array in C

I have two arrays, each initialised as follows:

struct file_descriptor *list1 = (struct file_descriptor *)malloc(4096 * sizeof(struct file_descriptor));

struct file_descriptor *list2 = (struct file_descriptor *)malloc(4096 * sizeof(struct file_descriptor));

Each array is essentially a list of struct file_descriptor pointers.

Now list1 already has a pointers in the array (that point to struct file_descriptor's), and I need to have the same indices in list2 point to the same objects that list1 pointers point to. So if list1[2] = some_object, I'm trying to point list2[2] = some_object as well.

So:

int i;
for(i = 3; i < 4096; i++) {
    struct file_descriptor *parent_fd = &list1[i]; // points to the struct that element at i points to in list1
    list2[i] = parent_fd;
}

At line 2 in the loop, this throws a: incompatible types when assigning to type ‘struct file_descriptor’ from type ‘struct file_descriptor *’ error.

Obviously, that's because list2[i] is the actual struct file_descriptor, and not the pointer in the list2 which points to that struct. Changing it to &list2[i] = parent_fd; also throws an error because &list2[i] is not an lvalue.

How would I dereference the pointer in the array at index i, so that list2[i] points to *list1[i]

Upvotes: 0

Views: 552

Answers (1)

R&#252;ppell&#39;s Vulture
R&#252;ppell&#39;s Vulture

Reputation: 3663

list1 and list2 are of type struct file_descriptor ** not struct file_descriptor *.

Further, don't declare a struct file_descriptor pointer or variable inside a loop.

Anyways,if you want the pointers in list2 to point to the same structure variables as the respective pointers in list1 do,then it's fairly simple.You do it as:

 for(i=0;i<=4096;i++)
 list2[i]=list1[i];

Further, if you intend your arrays to be arrays of pointers to the the structure variables, instead of arrays of structure variables, then you should use the following:

   malloc(4096 * sizeof(struct file_descriptor*));  //note the second *

Edit Your line list2[i] = parent_fd; shows type mismatch because you have declared list2 as of type struct file_descriptor* and hence list2[i] is of type struct file_descriptor, while parent_fd is of type struct file_descriptor*.In simple words,the left side of the = is a struct variable while the right side is a pointer to a struct variable. Like I said, you should declare list1 and list2 as type struct file_descriptor** not struct file_descriptor*.That should deal with the warning you encountered.

Upvotes: 1

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