Re: Is it a good practice to call the destructor explicitly and use
placement new(this) in assignment operators?
Paavo Helde wrote:
Pete Becker <pete@versatilecoding.com> wrote in news:-
eudnfqVh_ZQ3bjWnZ2dnUVZ_t-onZ2d@giganews.com:
Paavo Helde wrote:
Pete Becker <pete@versatilecoding.com> wrote in
news:mt6dnQGnBLwuirjWnZ2dnUVZ_ohi4p2d@giganews.com:
Paavo Helde wrote:
Michael Tsang <miklcct@gmail.com> wrote in
news:hg2tuh$bmb$1@news.eternal-september.org:
Pete Becker wrote:
No. Think about what happens when someone derives from this class
and
writes a normal assignment operator.
Sorry I forgot to make my assignment operator virtual. Now consider
the following code:
class Foo {
public:
// default constructor
Foo();
// l-value copy constructor
Foo(const Foo &);
// destructor
virtual ~Foo();
// assignment operator
virtual Foo &operator=(const Foo &x) {
if(this != &x) {
this->~Foo();
new(this) Foo(x);
This slices any derived class object to Foo.
It doesn't even do that. Slicing is well defined:
Foo f = Bar(); // slices the Bar object; f is a valid Foo object
That placement new constructs a Foo object where a Bar object used to
exist, resulting in undefined behavior.
I cannot quite see how this is UB by itself. Consider:
void* p = malloc(sizeof(Bar));
Bar* b = new (p) Bar();
b->~Bar();
Foo* f = new (p) Foo();
Here also a Foo object is constructed where Bar was. Is this UB too?
That's not inside operator=.
I presume this means no UB in the last example.
Then, how the operator= is any different? Clearly, just being a member
function should not affect anything. For example, several people (incl.
James Kanze) advocate using "delete this;" from inside a member function.
This clearly invalidates the object, but as long as it is not accessed
any more, there should be no UB.
operator= isn't usually called on objects that are going to be thrown
away immediately.
--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of
"The Standard C++ Library Extensions: a Tutorial and Reference"
(www.petebecker.com/tr1book)
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