Why can't we declare a variable of type void?

Question

I\'m looking for a formal explanation of that fact in the Standard. I\'ve found what 3.9.1/9 says and trying to give an explanation used that section.

S

Answer 1

Well - I really don't see the rationale behind this. It's going to be great if this way we can declare a variable with unknown type. Something like 'void *' and arrays of unknown size. Imagine code like this:

#include 
#include 

using namespace std;

extern void f;

int main()
{
    cout << (int &)f << endl; //cout 'f' as it was integer
}

struct {
    int a;
    double b;
} f{};

You can now actually do something similar with arrays:

#include 
#include 

using namespace std;

struct Foo;

extern int arr[];

int main()
{
    cout << arr[2] << endl;
}

int arr[4]{};

Life example.