consider class a:
class { public: int x; int y; } now if write a *a = new a() a pointer element of class a. however, if write a *a = new a[5] a[0] not pointer instance of class.
i expecting a[n] behave pointer , operations a[0]->x hold valid.
what flaw in understanding?
when a pointer, syntax a[n] is, definition, equivalent (*(a + n)). dereference included in expression. difference between new a() , new a[n] in what's allocated (a single element vs. multiple elements). however, in both cases, pointer single element returned (the element in case of new a(), , first element in case of new a[n]). , syntax can used on pointer same in both cases. note following:
a* = new a(); a[0].x = 10; this legal (albeit uncommon) way access single object allocated expression new a(). likewise:
a* = new a[n]; a->x = 10; that legal way access first element of array allocated expression new a[n].
