Zero size arrays in gcc c++ compiler - c++

Question

What will be the output of following c++ code snippet.
How we can assign a value to a variable which occupies no memory (0 bytes).Also doing pointer arithmetic gives false results(implied)
int main()
{
int arr[0];
arr[1]=1;
cout<<arr[1]<<endl;
cout<<sizeof(arr)<<endl;
int *p=arr;
int *q=p+1;
cout<<p-q;
return 0;
}
Sample execution ==> https://code.hackerearth.com/f8d7b1G

Answer 1

Well, actually, your code causes undefined behaviour.
Let's see how it works.
int a, b;
int c[2];
This will look like (imagine this as a stack inside the function "main"):
/***
+--------------------+ <- 0xXXXXXXXX (initial address)
| a |
+--------------------+ <- 0xXXXXXXXX + sizeof int (initial address + size of variable a) == &a
| b |
+--------------------+ <- ... + sizeof int (... + size of variable b) == &b
| c[1] |
+--------------------+
| c[0] |
+--------------------+ <- ... + number of bytes that are being allocated by the array (it's the pointer to the array)
== c or &c[0]
+--------------------+
| c (pointer) |
+--------------------+
// remember that compiler don't have to allocate it like this, it's just an example
***/
When you are trying to do something like this:
c[0] = 1;
it's being the same as:
*(c + 0) = 1;
because variable "c" contains the pointer to the array. It also explains why does a compiler allocate it reversely on the stack (because if we want to access a specific member of an array, we have to write something i did a little bit upper — *(c + N) = 1;, because STACK GROWS DOWN. If you still confused with it, write it in the comments below, i'll explain.
So, if you write something like this
int a, b;
int c[0];
It looks like:
/***
+--------------------+ <- 0xXXXXXXXX (initial address)
| a |
+--------------------+ <- 0xXXXXXXXX + sizeof int (initial address + size of variable a) == &a
| b |
+--------------------+ <- ... + sizeof int AND
+ number of bytes that are being allocated by the
array.
When the number of bytes is ZERO,
it equals to the address of the variable "b",
because &b + 0 == &b
+--------------------+
| c (pointer) |
+--------------------+
***/
So, we could say that what you are doing is:
*(&b + 1) = 1;
THIS IS VERY UNSAFE. Imagine there was no variable "b" or variable "a", you would just be managing something you shouldn't.
That's why zero-sized arrays are not allowed in C++ standart.