I'm dang certain that this code ought to be illegal, as it clearly won't work, but it seems to be allowed by the C++0x FCD.
class X { /* ... */};
void* raw = malloc(sizeof (X));
X* p = new (raw) X(); // according to the standard, the RHS is a placement-new expression
::operator delete(p); // definitely wrong, per litb's answer
delete p; // legal? I hope not
Maybe one of you language lawyers can explain how the standard forbids this.
There's also an array form:
class X { /* ... */};
void* raw = malloc(sizeof (X));
X* p = new (raw) X[1]; // according to the standard, the RHS is a placement-new expression
::operator delete[](p); // definitely wrong, per litb's answer
delete [] p; // legal? I hope not
This is the closest question I was able to find.
EDIT: I'm just not buying the argument that the standard's language restricting arguments to function void ::operator delete(void*)
apply in any meaningful way to the operand of delete
in a delete-expression. At best, the connection between the two is extremely tenuous, and a number of expressions are allowed as operands to delete
which are not valid to pass to void ::operator delete(void*)
. For example:
struct A
{
virtual ~A() {}
};
struct B1 : virtual A {};
struct B2 : virtual A {};
struct B3 : virtual A {};
struct D : virtual B1, virtual B2, virtual B3 {};
struct E : virtual B3, virtual D {};
int main( void )
{
B3* p = new E();
void* raw = malloc(sizeof (D));
B3* p2 = new (raw) D();
::operator delete(p); // definitely UB
delete p; // definitely legal
::operator delete(p2); // definitely UB
delete p2; // ???
return 0;
}
I hope this shows that whether a pointer may be passed to void operator delete(void*)
has no bearing on whether that same pointer may be used as the operand of delete
.
See Question&Answers more detail:
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