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Rollup merge of #28797 - steveklabnik:rebase_for_gankro, r=alexcrichton
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This is a rebase of #27902, since @gankro  is busy ❤️
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steveklabnik committed Oct 1, 2015
2 parents 979c595 + c7527a1 commit db27dca
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75 changes: 75 additions & 0 deletions src/libcollections/vec.rs
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Expand Up @@ -148,6 +148,81 @@ use super::range::RangeArgument;
/// if the vector's length is increased to 11, it will have to reallocate, which
/// can be slow. For this reason, it is recommended to use `Vec::with_capacity`
/// whenever possible to specify how big the vector is expected to get.
///
/// # Guarantees
///
/// Due to its incredibly fundamental nature, Vec makes a lot of guarantees
/// about its design. This ensures that it's as low-overhead as possible in
/// the general case, and can be correctly manipulated in primitive ways
/// by unsafe code. Note that these guarantees refer to an unqualified `Vec<T>`.
/// If additional type parameters are added (e.g. to support custom allocators),
/// overriding their defaults may change the behavior.
///
/// Most fundamentally, Vec is and always will be a (pointer, capacity, length)
/// triplet. No more, no less. The order of these fields is completely
/// unspecified, and you should use the appropriate methods to modify these.
/// The pointer will never be null, so this type is null-pointer-optimized.
///
/// However, the pointer may not actually point to allocated memory. In particular,
/// if you construct a Vec with capacity 0 via `Vec::new()`, `vec![]`,
/// `Vec::with_capacity(0)`, or by calling `shrink_to_fit()` on an empty Vec, it
/// will not allocate memory. Similarly, if you store zero-sized types inside
/// a Vec, it will not allocate space for them. *Note that in this case the
/// Vec may not report a `capacity()` of 0*. Vec will allocate if and only
/// if `mem::size_of::<T>() * capacity() > 0`. In general, Vec's allocation
/// details are subtle enough that it is strongly recommended that you only
/// free memory allocated by a Vec by creating a new Vec and dropping it.
///
/// If a Vec *has* allocated memory, then the memory it points to is on the heap
/// (as defined by the allocator Rust is configured to use by default), and its
/// pointer points to `len()` initialized elements in order (what you would see
/// if you coerced it to a slice), followed by `capacity() - len()` logically
/// uninitialized elements.
///
/// Vec will never perform a "small optimization" where elements are actually
/// stored on the stack for two reasons:
///
/// * It would make it more difficult for unsafe code to correctly manipulate
/// a Vec. The contents of a Vec wouldn't have a stable address if it were
/// only moved, and it would be more difficult to determine if a Vec had
/// actually allocated memory.
///
/// * It would penalize the general case, incurring an additional branch
/// on every access.
///
/// Vec will never automatically shrink itself, even if completely empty. This
/// ensures no unnecessary allocations or deallocations occur. Emptying a Vec
/// and then filling it back up to the same `len()` should incur no calls to
/// the allocator. If you wish to free up unused memory, use `shrink_to_fit`.
///
/// `push` and `insert` will never (re)allocate if the reported capacity is
/// sufficient. `push` and `insert` *will* (re)allocate if `len() == capacity()`.
/// That is, the reported capacity is completely accurate, and can be relied on.
/// It can even be used to manually free the memory allocated by a Vec if
/// desired. Bulk insertion methods *may* reallocate, even when not necessary.
///
/// Vec does not guarantee any particular growth strategy when reallocating
/// when full, nor when `reserve` is called. The current strategy is basic
/// and it may prove desirable to use a non-constant growth factor. Whatever
/// strategy is used will of course guarantee `O(1)` amortized `push`.
///
/// `vec![x; n]`, `vec![a, b, c, d]`, and `Vec::with_capacity(n)`, will all
/// produce a Vec with exactly the requested capacity. If `len() == capacity()`,
/// (as is the case for the `vec!` macro), then a `Vec<T>` can be converted
/// to and from a `Box<[T]>` without reallocating or moving the elements.
///
/// Vec will not specifically overwrite any data that is removed from it,
/// but also won't specifically preserve it. Its uninitialized memory is
/// scratch space that it may use however it wants. It will generally just do
/// whatever is most efficient or otherwise easy to implement. Do not rely on
/// removed data to be erased for security purposes. Even if you drop a Vec, its
/// buffer may simply be reused by another Vec. Even if you zero a Vec's memory
/// first, that may not actually happen because the optimizer does not consider
/// this a side-effect that must be preserved.
///
/// Vec does not currently guarantee the order in which elements are dropped
/// (the order has changed in the past, and may change again).
///
#[unsafe_no_drop_flag]
#[stable(feature = "rust1", since = "1.0.0")]
pub struct Vec<T> {
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