On Jun 28, 6:13 am, Guillermo Schwarz <guillermo.schw...@gmail.com>
wrote:
On Jun 27, 12:46 pm, Diego Martins <jose.di...@gmail.com> wrote:
On Jun 26, 7:55 am, dasjotre <dasjo...@googlemail.com> wrote:
struct Base
{
virtual ~Base()=0{}
this is a syntax error
it should be
struct Base
{
virtual ~Base()=0;};
...
Base::~Base() {}
Good point. Also why destructor can be non virtual? Is there any use
for that? I have never seen a real world scenario where destructors
shoudl not be virtual.
Really? What about structures which are shared with C? Putting
a vptr into the structure isn't going to please the C compiler
any. For that matter, making the destructor virtual for
something like complex<float> could double the size of the
object. An important consideration for people working with
large arrays of the thing. In general, for classes with value
semantics, there's no reason for the destructor to be virtual,
and every reason for it not to be.
Once a class has virtual functions, of course, the question
changes. The vptr is already there, so making the destructor
virtual doesn't change anything in this respect. And the
presence of virtual functions means that there is a very high
probability that inheritance will be involved somewhere, and
that a virtual destructor will be needed. On the other hand,
making the virtuality of the destructor depend on whether other
virtual functions are present or not isn't very orthogonal, to
say the least, and is another complication. And IMHO, C++
already has enough special cases, and doesn't really need
another one.