Reserve(n) for unordered containers reserves for n-1 elements.

[Daniel James]

Hello,

When calling 'x.reserve(n)' for an unordered container I'd expect to
be able to insert n elements without invalidating iterators. But as
the standard is written, the guarantee only holds for n-1 elements.
For a container with max_load_factor of 1, reserve(n) is equivalent to
rehash(ceil(n/1)), which is rehash(n). rehash(n) requires that the
bucket count is >= n, so it can be n. The rule is that insert shall
not affect the validity of iterators if (N + n) < z * B. But for this
case the two sides of the equation are equal, so insert can affect the
validity of iterators. Is that a mistake?

Daniel


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[Daniel Krügler]

Am 03.05.2012 20:06, schrieb Daniel James:
> When calling 'x.reserve(n)' for an unordered container I'd expect to
> be able to insert n elements without invalidating iterators. But as
> the standard is written, the guarantee only holds for n-1 elements.
> For a container with max_load_factor of 1, reserve(n) is equivalent to
> rehash(ceil(n/1)), which is rehash(n). rehash(n) requires that the
> bucket count is>= n, so it can be n. The rule is that insert shall
> not affect the validity of iterators if (N + n)< z * B. But for this
> case the two sides of the equation are equal, so insert can affect the
> validity of iterators. Is that a mistake?

I tend to say that this is indeed a mistake. Please submit a library
defect to the email address referenced in the reply-to field of

http://www.open-std.org/jtc1/sc22/wg21/docs/lwg-active.html

Thanks& Greetings from Bremen,

Daniel Krügler