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Is the following code well-formed and portable?

Value value;
Container<Value> c;

c.push_back(value);
c.erase(--c.end());

Container can be std::string, std::vector...

On Apr 8, 11:46*am, Pete Becker <p...@versatilecoding.com> wrote:
> It's not quite that simple. If the iterator is a class type you can
> decrement it. So if the iterator isn't a class type, the code is
> ill-formed; if the iterator is a class type, the code works fine. But
> it's not portable.

Using rbegin() instead of --end() should fix that.

--Jonathan

On Apr 8, 4:46 pm, Pete Becker <p...@versatilecoding.com> wrote:
> Victor Bazarov wrote:
> > keith wrote:
> >> Is the following code well-formed and portable?

> >> Value value;
> >> Container<Value> c;

> >> c.push_back(value);
> >> c.erase(--c.end());

> >> Container can be std::string, std::vector...

> > First thing that came to my mind: c.end() returns a
> > *temporary* which isn't an lvalue. Decrement requires an
> > lvalue, hence the code that has (--c.end()) is ill-formed.
> > Ill-formed program cannot be portable. So, the answer to
> > your question is "No".

> It's not quite that simple. If the iterator is a class type you can
> decrement it.

Maybe. If the iterator is a class type and the increment
operator is a member, you can decrement it. If the increment
operator is a free function, taking the iterator as a non-const
reference, you can't. (I don't know of any implementations
which do it that way, but I don't know why. It's the way I'd do
it.)

--
James Kanze