Re: std::vector<T>::emplace and exception safety

From:
Fuz <srobb@reflectionsinteractive.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Sat, 18 Aug 2007 20:23:29 CST
Message-ID:
<1187480211.913517.21760@d55g2000hsg.googlegroups.com>
On Aug 17, 9:18 pm, David Abrahams <d...@boost-consulting.com> wrote:

This was mentioned earlier by ymett. But now that I've checked the
Standard, I don't think you can.


Trust me, you can.


I'll need a little more to go on than blind faith, I'm afraid.

If no reallocation happens, all iterators and references before the
insertion point remains valid.


So what?


There's no need to be rude.

I'm saying that you can't maintain valid iterators and references if
you're clearing the container. All the Standard says is: "If no
reallocation happens, all the iterators and references before the
insertion point remain valid.". There's no extra qualification along
the lines of: 'unless an exception is thrown'. Therefore the
following should be possible:

struct type
{
     type(int); // could throw
};

std::vector<type> v;
v.reserve(10);
v.push_back(1);
v.push_back(3);
type& ref = v.front();
try
{
     v.emplace(v.begin() + 1, 2);
}
catch (const type_exception&)
{
     // can use ref here - no rellocation has taken place, so it should
be valid
}

You're saying that it's valid for an implementation to clear if
emplace (or any other type of insert) has failed. The Standard says
that ref must be valid, which prevents that possibility.

You're adamant that clearing is allowed, but I can't see how that line
in the Standard can be interpreted any other way. So I'm politely
asking you to explain how it's possible.

Yes. And then you can destroy the rest of the elements if you want.
Or permute them, for that matter.


Permute them, reassign them, anything... except destroying them.

It does "seem" wrong at first, but eventually it becomes obvious that
allowing data to be lost is exactly right. The problem is that:

<snip>


Ok, I'm convinced. My main point was not that the strong guarantee
was required, but just some guarantee that the original contents of
the container were in there *somewhere*. But you're right, it's too
much for implementations to worry about for a not-actually-very-useful
case.

In fact, this is good for std::vector<T>::emplace. It means that it
can always placement-new can always be used, without the generation of
a temporary, which was my main worry.

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