c，将void *指向的值赋给void * array

Question

I am trying to assign a value pointed at by a void * in a void * array. Here is what I have so far:

23 void queue_enqueue(void *q, void *item) {
24 int len = 0 ;
25 
26 len = sizeof(q) ;
27 q[len-1] = 
28     item ;
29     
30 return ;
31 
32 }

With this compiler error:

myqueue.c:27: warning: dereferencing ‘void *’ pointer
myqueue.c:28: error: invalid use of void expression

I've been searching for an answer for about a hour now, but I haven't seen an implementation similar to what I have been given here. In case you were wondering, yes this is homework and the prototype was given by the instructor.

Because they are passed as void * I have been unable to use the gcc typeof operator.

I'm okay with the warning. I know it's just gcc telling me that "be careful we're not going to check this for you", but I can't get rid of the error on line 28.

Thank you for your help in advance.

Answer 1

That's not an array of void* s, it's an array of void (which doesn't really exist), and dereferencing it gets you a void , which is "not a real type", so you can't assign to it.

If you're trying to work with an array of void* , change it to

void queue_enqueue(void** q, void* item)

or

void queue_enqueue(void* q[], void* item) // same as above

Also realise that your program will do

q[sizeof(void**) - 1] = item;

every time. sizeof gives you the size of the pointer type, not the array. You have to pass the length of an array as a separate argument if you want to know how long an array is.

Answer 2

你需要一个void*数组：

void queue_enqueue(void *q[], void *item)

Answer 3

void is a type with no size. Thus an array of void cannot exist. (because an array defines the amount of units and then jumps using the sizeof() operator.

you would have to cast the void* to something*, something having the same size of what you actually have stored there, so say you have type T, which is of same SIZE( I'm not saying same thing ),

T* q2 = (T*)q;
T* item2 = (T*)item;
q2[len-1] = &item;

This will work because C passes pointers by reference.
since T will be a real type the compiler lets you do it
since T is same size as your actual stored values, you will wind up with the correct value storage location.

note that void* as an array, will contain a pointer to an array of type void. void* as a single value will contain a single value of void.