Void pointers in place of pointers of type int

Question

I am a beginner in C/C++ and i am trying to learn the pointers.

Here is my Code to create the array of pointers with each element in the pointer array, pointing to the element in the data array:

#include <iostream>
using namespace std;

//Pointers reference article
//https://www.programiz.com/cpp-programming/pointers-arrays

/* Array of pointers */
const int MAX = 5;
int main(){
    int arr[MAX] = {1,2,3,4,5};
    int* ptr[MAX];

    cout << "Create the handle of each element in data array to the ptr array: " << endl;
    for (int i=0; i<sizeof(arr)/sizeof(arr[0]);i++)
    {
        ptr[i] = &arr[i];
        cout<<"ptr["<<i<<"] = " << ptr[i] << endl;
    }

    cout << "Display the contents of array using 1:1 ptr array:"<< endl;
    for (int i=0; i<sizeof(arr)/sizeof(arr[0]);i++)
        cout<<"arr["<<i<<"] = " << *ptr[i] << endl;

    system ("pause");
    return 0;
}

The above program works as expected. But, if i change the pointer type from int to void during pointer declaration, i.e from int* ptr[MAX]; to void* ptr[MAX];

I have this error: cpp(22): error C2100: illegal indirection

Line 22: cout<<"arr["<<i<<"] = " << *ptr[i] << endl

Can someone please educate me on this error. Thanks in advance.

If you dereference a `void *`, you just get a `void`. I'm not really sure what you would expect that to do. — Dolda2000, Jan 15 '17 at 04:04
There is no language C/C++. One the distinct languages C and C++. This code is clearly not C. — too honest for this site, Jan 15 '17 at 04:04
The whole point of a `void` pointer is that it can point at (almost) anything, regardless of type, so the type of what it points at is unknown. To dereference a pointer (as in `*ptr[i]`, where `ptr[i]` is a pointer) the type of what it points at must be known. — Peter, Jan 15 '17 at 04:04
Not related to the question, but I recommend declaring pointer variables as `int *x;` instead of `int* x;`, because the `*` modifies the declarator rather than the type. For instance, `int* a, b;` gives the impression that both `a` and `b` are `int` pointers, when in fact only `a` is while `b` is just an `int`. `int *a, b;` does somewhat better at avoiding that misunderstanding. (Though arguably `int *a; int b;` is the less confusing construction.) — Dolda2000, Jan 15 '17 at 04:08
@Dolda2000 Whatever declare more than one variable in a statement is also a bad practice so I'm not agree with your argument. — Stargateur, Jan 15 '17 at 04:27
@Dolda2000 I would recommend the opposite because in `C++` the `*` is part of the type in a declaration (C++ is type-centric). This is what Bjarne Stroustrup recommends. — Galik, Jan 15 '17 at 04:45
@Galik: I just tried, and using G++, `int* a, b` certainly declared `b` as an `int`, not an `int *`. — Dolda2000, Jan 15 '17 at 04:47
@Dolda2000 Not the place to debate it, its personal preference. But no one should be using that construct for several reasons. Even so the `*` is still part of the *type* information. This is what Bjarne suggests and that is what is used in the `C++` Standard documents. The old "expression centric" view is more favored by `C` programmers from what I can see. Also `int* p;` is more intuitive and easier to comprehend for beginners. Plus the "expression centric" view doesn't make sense for *references*. — Galik, Jan 15 '17 at 04:56
@Galik technically the language grammar defines a declaration as a type specifier followed by a list of declarators, and the way the grammar is written, the `*` is part of the declarator, not the type specifier. But I don't think that has anything to do with the style decision. I have seen people advocate `int *p` but `int& q`. — M.M, Jan 15 '17 at 05:01
BTW i wonder if there will ever be a question mentioning a pointer without somebody coming along to impose their personal style in comments — M.M, Jan 15 '17 at 05:02

score 4 · Accepted Answer · answered Jan 15 '17 at 04:06

int* ptr[MAX];

ptr is an array of pointer to int.

When you change to

void* ptr[MAX];

then ptr is an array of pointer to void. That does not cause an error on the the first cout

cout<<"ptr["<<i<<"] = " << ptr[i] << endl;  // ok - printing the address

But it's an error on the second:

cout<<"arr["<<i<<"] = " << *ptr[i] << endl;  // error - dereferencing void pointer

You cannot dereference a void pointer - that's what the error about. A pointer must be of specific type to be dereferenced.

score 2 · Answer 2 · answered Jan 15 '17 at 04:05

You need to ask yourself one question - What does void mean?

So you have a void pointer - this means it points in to the void

That can be anything - Any int, A structure, An object, A float....

Get the picture

Then the compiler is trying to de-reference it - So it holds up its hands and say - I have not got a clue.

Either case it - or better still Avoid void pointers

Void pointers in place of pointers of type int

2 Answers2