c++ - Set value of random memory address -

i understand isn't practice following. however, curious if possible?

x = 1; cout << *(&x + 1) << endl; // prints value @ next memory address. (&x + 1) = 0; // doesn't work.  // want set value of next memory address. 

the error says:

error: lvalue required left operand of assignment

is there can change make work, though understand isn't useful.

the error due trying assign temporary address.

(&x + 1) = 0; 

what meant was:

*(&x + 1) = 0; 

the results of expression depends on type of x. if x of type int, result in undefined behaviour assigning memory have not been allocated.

however, using type following result in defined behaviour example.

struct foo {     foo& operator=(int i) { x[0] = i; return *this; }     int* operator&() { return x.data(); }     std::array<int, 2> x{{0}}; };  int main() {     foo x;     x = 1;     std::cout << *(&x + 1) << std::endl; // equivalent 'x.x[1]' '0'.     *(&x + 1) = 0; // ok. } 

in c++, undefined behaviour access memory have not been allocated. results of such operation can result in pretty as, e.g., execution of random instruction laying around in deallocated memory.