When Does Move Constructor get called?

A move constructor is called:

• when an object initializer is std::move(something)
• when an object initializer is std::forward<T>(something) and T is not an lvalue reference type (useful in template programming for "perfect forwarding")
• when an object initializer is a temporary and the compiler doesn't eliminate the copy/move entirely
• when returning a function-local class object by value and the compiler doesn't eliminate the copy/move entirely
• when throwing a function-local class object and the compiler doesn't eliminate the copy/move entirely

This is not a complete list. Note that an "object initializer" can be a function argument, if the parameter has a class type (not reference).

a RetByValue() {
    a obj;
    return obj; // Might call move ctor, or no ctor.
}

void TakeByValue(a);

int main() {
    a a1;
    a a2 = a1; // copy ctor
    a a3 = std::move(a1); // move ctor

    TakeByValue(std::move(a2)); // Might call move ctor, or no ctor.

    a a4 = RetByValue(); // Might call move ctor, or no ctor.

    a1 = RetByValue(); // Calls move assignment, a::operator=(a&&)
}

First of all, your copy constructor is broken. Both the copied from and copied to objects will point to the same Array and will both try to delete[] it when they go out of scope, resulting in undefined behavior. To fix it, make a copy of the array.

a::a(const a& Old): Array(new int[5])
{
  for( size_t i = 0; i < 5; ++i ) {
    Array[i] = Old.Array[i];
  }
}

Now, move assignment is not being performed as you want it to be, because both assignment statements are assigning from lvalues, instead of using rvalues. For moves to be performed, you must be moving from an rvalue, or it must be a context where an lvalue can be considered to be an rvalue (such as the return statement of a function).

To get the desired effect use std::move to create an rvalue reference.

A=C;              // A will now contain a copy of C
B=std::move(C);   // Calls the move assignment operator