realloc problem on compilation

Question

hello. I am trying to implement queues in C.
I wrote this as queue.h:
```
typedef struct {
void * arr[1];
int len, head, tail;
} queue;
```
and, in test.c:
```
#include <stdio.h>
#include <stdlib.h>
#include "queue.h"

queue spawn(int len){
queue q;
q.arr = (void *) realloc(q.arr, len * sizeof(void *)); /* = (void *) realloc(...);*/
for(int i = 0; i < len; i++){
q.arr[i] = NULL;
}
q.len = len, q.head = 0, q.tail = 0;
return q;
}
```
But, when I runned gcc, i got:
```
error: assignment to expression with array type
q.arr = (void *) realloc(q.arr, len * sizeof(void *));
```
I began reading this post:
https://stackoverflow.com/questions/37225244/error-assignment-to-expression-with-array-type-error-when-i-assign-a-struct-f
But it doesn't answer my problem. I tthink i did everything right tho, as I saw in the C course and here:
https://www.tutorialspoint.com/c_standard_library/c_function_realloc.htm
but it seems that I didn't. What must I do?

Answer 1

I tried changing the following
q.arr =
To
q.arr[0] = Or *q.arr =
And now gcc compiles that piece of code without errors.

q.arr its just an address it cant hold any value
q.arr[0] is a pointer variable which can hold an address that why it can recieve the address from the realloc function.
*q.arr targets the content of that address which is a pointer variable.

*q.arr is same as q.arr[0]

That array of pointers + structs + memory allocation is still tricky to me I need to study and practice more, it requires solid foundation. 😥
I hope some advanced C coders take a look at your problem. Good luck mate!

Answer 2

@arturop I'm implementing the queue as an array of void pointers. But with what you suggest me, the code would be changed: my aim is here to give the queue a non-default size, while what you do is something completely different. Thanks tho for answering!

Answer 3

Gwlanbzh 😅✌