2020-12-04 13:47:15 +00:00
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use std::alloc::{Allocator, Global, Layout, System};
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alloc_system: don’t assume MIN_ALIGN for small sizes, fix #45955
The GNU C library (glibc) is documented to always allocate with an alignment
of at least 8 or 16 bytes, on 32-bit or 64-bit platforms:
https://www.gnu.org/software/libc/manual/html_node/Aligned-Memory-Blocks.html
This matches our use of `MIN_ALIGN` before this commit.
However, even when libc is glibc, the program might be linked
with another allocator that redefines the `malloc` symbol and friends.
(The `alloc_jemalloc` crate does, in some cases.)
So `alloc_system` doesn’t know which allocator it calls,
and needs to be conservative in assumptions it makes.
The C standard says:
https://port70.net/%7Ensz/c/c11/n1570.html#7.22.3
> The pointer returned if the allocation succeeds is suitably aligned
> so that it may be assigned to a pointer to any type of object
> with a fundamental alignment requirement
https://port70.net/~nsz/c/c11/n1570.html#6.2.8p2
> A fundamental alignment is represented by an alignment less than
> or equal to the greatest alignment supported by the implementation
> in all contexts, which is equal to `_Alignof (max_align_t)`.
`_Alignof (max_align_t)` depends on the ABI and doesn’t seem to have
a clear definition, but it seems to match our `MIN_ALIGN` in practice.
However, the size of objects is rounded up to the next multiple
of their alignment (since that size is also the stride used in arrays).
Conversely, the alignment of a non-zero-size object is at most its size.
So for example it seems ot be legal for `malloc(8)` to return a pointer
that’s only 8-bytes-aligned, even if `_Alignof (max_align_t)` is 16.
2017-11-20 14:30:04 +00:00
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2019-07-02 10:56:51 +00:00
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/// Issue #45955 and #62251.
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alloc_system: don’t assume MIN_ALIGN for small sizes, fix #45955
The GNU C library (glibc) is documented to always allocate with an alignment
of at least 8 or 16 bytes, on 32-bit or 64-bit platforms:
https://www.gnu.org/software/libc/manual/html_node/Aligned-Memory-Blocks.html
This matches our use of `MIN_ALIGN` before this commit.
However, even when libc is glibc, the program might be linked
with another allocator that redefines the `malloc` symbol and friends.
(The `alloc_jemalloc` crate does, in some cases.)
So `alloc_system` doesn’t know which allocator it calls,
and needs to be conservative in assumptions it makes.
The C standard says:
https://port70.net/%7Ensz/c/c11/n1570.html#7.22.3
> The pointer returned if the allocation succeeds is suitably aligned
> so that it may be assigned to a pointer to any type of object
> with a fundamental alignment requirement
https://port70.net/~nsz/c/c11/n1570.html#6.2.8p2
> A fundamental alignment is represented by an alignment less than
> or equal to the greatest alignment supported by the implementation
> in all contexts, which is equal to `_Alignof (max_align_t)`.
`_Alignof (max_align_t)` depends on the ABI and doesn’t seem to have
a clear definition, but it seems to match our `MIN_ALIGN` in practice.
However, the size of objects is rounded up to the next multiple
of their alignment (since that size is also the stride used in arrays).
Conversely, the alignment of a non-zero-size object is at most its size.
So for example it seems ot be legal for `malloc(8)` to return a pointer
that’s only 8-bytes-aligned, even if `_Alignof (max_align_t)` is 16.
2017-11-20 14:30:04 +00:00
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#[test]
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fn alloc_system_overaligned_request() {
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check_overalign_requests(System)
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}
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2017-11-20 14:42:34 +00:00
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#[test]
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fn std_heap_overaligned_request() {
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2018-04-03 19:15:06 +00:00
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check_overalign_requests(Global)
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2017-11-20 14:42:34 +00:00
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}
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2020-12-04 13:47:15 +00:00
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fn check_overalign_requests<T: Allocator>(allocator: T) {
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2019-07-02 10:56:51 +00:00
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for &align in &[4, 8, 16, 32] {
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// less than and bigger than `MIN_ALIGN`
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for &size in &[align / 2, align - 1] {
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// size less than alignment
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let iterations = 128;
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unsafe {
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let pointers: Vec<_> = (0..iterations)
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.map(|_| {
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2020-12-04 13:47:15 +00:00
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allocator.allocate(Layout::from_size_align(size, align).unwrap()).unwrap()
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2019-07-02 10:56:51 +00:00
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})
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.collect();
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for &ptr in &pointers {
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assert_eq!(
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2020-08-04 16:03:34 +00:00
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(ptr.as_non_null_ptr().as_ptr() as usize) % align,
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2019-07-02 10:56:51 +00:00
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0,
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"Got a pointer less aligned than requested"
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)
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}
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alloc_system: don’t assume MIN_ALIGN for small sizes, fix #45955
The GNU C library (glibc) is documented to always allocate with an alignment
of at least 8 or 16 bytes, on 32-bit or 64-bit platforms:
https://www.gnu.org/software/libc/manual/html_node/Aligned-Memory-Blocks.html
This matches our use of `MIN_ALIGN` before this commit.
However, even when libc is glibc, the program might be linked
with another allocator that redefines the `malloc` symbol and friends.
(The `alloc_jemalloc` crate does, in some cases.)
So `alloc_system` doesn’t know which allocator it calls,
and needs to be conservative in assumptions it makes.
The C standard says:
https://port70.net/%7Ensz/c/c11/n1570.html#7.22.3
> The pointer returned if the allocation succeeds is suitably aligned
> so that it may be assigned to a pointer to any type of object
> with a fundamental alignment requirement
https://port70.net/~nsz/c/c11/n1570.html#6.2.8p2
> A fundamental alignment is represented by an alignment less than
> or equal to the greatest alignment supported by the implementation
> in all contexts, which is equal to `_Alignof (max_align_t)`.
`_Alignof (max_align_t)` depends on the ABI and doesn’t seem to have
a clear definition, but it seems to match our `MIN_ALIGN` in practice.
However, the size of objects is rounded up to the next multiple
of their alignment (since that size is also the stride used in arrays).
Conversely, the alignment of a non-zero-size object is at most its size.
So for example it seems ot be legal for `malloc(8)` to return a pointer
that’s only 8-bytes-aligned, even if `_Alignof (max_align_t)` is 16.
2017-11-20 14:30:04 +00:00
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2019-07-02 10:56:51 +00:00
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// Clean up
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for &ptr in &pointers {
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2020-12-04 13:47:15 +00:00
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allocator.deallocate(
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2020-08-04 16:03:34 +00:00
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ptr.as_non_null_ptr(),
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Layout::from_size_align(size, align).unwrap(),
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)
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2019-07-02 10:56:51 +00:00
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}
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}
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alloc_system: don’t assume MIN_ALIGN for small sizes, fix #45955
The GNU C library (glibc) is documented to always allocate with an alignment
of at least 8 or 16 bytes, on 32-bit or 64-bit platforms:
https://www.gnu.org/software/libc/manual/html_node/Aligned-Memory-Blocks.html
This matches our use of `MIN_ALIGN` before this commit.
However, even when libc is glibc, the program might be linked
with another allocator that redefines the `malloc` symbol and friends.
(The `alloc_jemalloc` crate does, in some cases.)
So `alloc_system` doesn’t know which allocator it calls,
and needs to be conservative in assumptions it makes.
The C standard says:
https://port70.net/%7Ensz/c/c11/n1570.html#7.22.3
> The pointer returned if the allocation succeeds is suitably aligned
> so that it may be assigned to a pointer to any type of object
> with a fundamental alignment requirement
https://port70.net/~nsz/c/c11/n1570.html#6.2.8p2
> A fundamental alignment is represented by an alignment less than
> or equal to the greatest alignment supported by the implementation
> in all contexts, which is equal to `_Alignof (max_align_t)`.
`_Alignof (max_align_t)` depends on the ABI and doesn’t seem to have
a clear definition, but it seems to match our `MIN_ALIGN` in practice.
However, the size of objects is rounded up to the next multiple
of their alignment (since that size is also the stride used in arrays).
Conversely, the alignment of a non-zero-size object is at most its size.
So for example it seems ot be legal for `malloc(8)` to return a pointer
that’s only 8-bytes-aligned, even if `_Alignof (max_align_t)` is 16.
2017-11-20 14:30:04 +00:00
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}
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}
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}
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