Re: best way to "delete" all objects in a std::vector.
In article
<3e5d46fb-2fcb-4ed8-9110-ee8f715c7b0c@d45g2000hsc.googlegroups.com>,
James Kanze <james.kanze@gmail.com> wrote:
On Jun 4, 6:08 pm, "jason.cipri...@gmail.com"
<jason.cipri...@gmail.com> wrote:
On Jun 4, 7:25 am, James Kanze <james.ka...@gmail.com> wrote:
On Jun 4, 5:43 am, "Daniel T." <danie...@earthlink.net> wrote:
Daniel Pitts <newsgroup.spamfil...@virtualinfinity.net> wrote:
I have std::vector<Base *> bases;
I'd like to do something like:
std::for_each(bases.begin(), bases.end(), operator delete);
Is it possible without writing an adapter? Is there a better
way? Is there an existing adapter?
From Stroustrup's book.
struct Delete_ptr {
template<class T> T* operator()(T* p) const { delete p; return 0; }
};
...
transform(s.begin(),s.end(),s.begin(),Delete_ptr());
[To the original poster: ignore this: it is from an
obsessional nitpicker, only for expert nitpickers.]
Note that formally, the above still has undefined behavior,
since it leaves a deleted pointer in the container for a (very)
short time. The correct way of doing this would be:
What has undefined behavior? If you are talking about the
transform over all, or the () operator, neither are undefined.
It is well- defined that it leaves an invalid pointer in the
container for a very short time. What's undefined is if
something attempts to dereference that pointer during that
time.
A pointer, after delete, may not even be read or copied (since
one of the effects of delete could be to render its value
"trapping").
The implementation I presented doesn't read or copy any pointers after
the delete. The object is deleted, then the pointer is assigned NULL. If
the code I presented is "technically undefined" then so is:
delete p;
p = NULL;
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