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

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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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Reserve(n) for unordered containers reserves for n-1 elements. Daniel James <dnljms@gmail.com> - 2012-05-03 11:06 -0700
  Re: Reserve(n) for unordered containers reserves for n-1 elements. Daniel Krügler<daniel.kruegler@googlemail.com> - 2012-05-04 11:24 -0700

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