Question

What is the meaning of

*(int *)0 = 0; 

It does compile successfully

Answer 1

It will usually cause an access violation at runtime. The following is done: first 0 is cast to an int * and that yields a null pointer. Then a value 0 is written to that address (null address) – that causes undefined behaviour, usually an access violation.

Effectively it is this code:

int* address = reinterpret_cast<int*>( 0 );
*address = 0;