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doc: explain why it is unsafe to construct Vec<u8> from Vec<u16> #65873
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Original file line number | Diff line number | Diff line change |
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@@ -411,7 +411,11 @@ impl<T> Vec<T> { | |
/// | ||
/// Violating these may cause problems like corrupting the allocator's | ||
/// internal data structures. For example it is **not** safe | ||
/// to build a `Vec<u8>` from a pointer to a C `char` array and a `size_t`. | ||
/// to build a `Vec<u8>` from a pointer to a C `char` array with length `size_t`. | ||
/// It's also not safe to build one from a `Vec<u16>` and its length, because | ||
/// the allocator cares about the alignment, and these two types have different | ||
/// alignments. The buffer was allocated with alignment 2 (for `u16`), but after | ||
/// turning it into a `Vec<u8>` it'll be deallocated with alignment 1. | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. This reads just a little off; I would say /// to build a `Vec<u8>` from a pointer to a C `char` array with length `size_t`.
/// It's also not safe to build one from a `Vec<u16>` and its length, because
/// the allocator cares about the alignment, and these two types have different
/// alignments. The buffer was allocated with alignment 2 (for `u16`), but after
/// turning it into a `Vec<u8>` it'll be deallocated with alignment 1.
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||
/// | ||
/// The ownership of `ptr` is effectively transferred to the | ||
/// `Vec<T>` which may then deallocate, reallocate or change the | ||
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