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Zitat

Also ist 0 gleichzusetzen mit einem NULL-Zeiger ?

int *pNull;

pNull = 0;
delete[] pNull; //passiert nichts

pNull = NULL;
delete[] pNull; //hier auch nicht

Zitat

das dritte wird nicht klappen, weil es in C++ keine automatische Umwandlung von void* zu anderen Zeigern gibt. Ansonsten hast du recht.

Zitat

In the second alternative (delete array), the value
of the operand of delete shall be the pointer value which resulted
from a previous array new-expression.18 ) If not, the behavior is unde-
fined. [Note: this means that the syntax of the delete-expression
must match the type of the object allocated by new, not the syntax of
the new-expression. ] [Note: a pointer to a const type can be the
operand of a delete-expression; it is not necessary to cast away the
constness (_expr.const.cast_) of the pointer expression before it is
_________________________
18 ) For non-zero-length arrays, this is the same as a pointer to the
first element of the array created by that new-expression. Zero-
length arrays do not have a first element.