`mimalloc` returns non-aligned memory

[emilk]

I had this code fail (on both Mac and Windows):

#[global_allocator]
static GLOBAL: AccountingAllocator<mimalloc::MiMalloc> =
    AccountingAllocator::new(mimalloc::MiMalloc);

#[repr(C, align(256))]
pub struct BigStruct { … }

fn collect() {
    assert_eq!(align_of::<BigStruct>(), 256);
    assert_eq!(size_of::<BigStruct>(), 256);
    let vec: Vec<T> = std::iter::once(BigStruct{…}).collect();
    dbg!(vec.as_ptr() as usize % std::mem::align_of::<T>()); // Sometimes prints 64
    let slice = vec.as_slice(); // panics in debug builds because of the unaligned memory
}

Granted, 256 is a very large alignment, but I expect either mimalloc to handle it, or to log a warning, or panic, or similar.
But I don't want it to quietly return unaligned memory.
I cannot find any docs saying there is a limit to the supported alignment.

The standard allocator never fails on the same code.

Maybe related

• mi_heap_realloc_aligned does not respect alignment microsoft/mimalloc#603

[nathaniel-daniel]

I'm assuming this is from https://github.com/rerun-io/rerun

Looks like you've pinned your mimalloc dep to 0.1.37. The alignment issues were fixed in 0.1.39. You can also potentially work around that by disabling the secure feature (disable default features), but I didn't extensively check to see if that fixed things.

Relevant Issue: #87

Also, that if std::mem::align_of::<T>() <= 64 || pretend_mimalloc_aligns_correctly isn't sufficient, I believe it can fail for an alignment of 64 as well.

[emilk]

Thank you @nathaniel-daniel !

I did run cargo update -p mimalloc and when it didn’t update I assumed we were using the latest version, not realizing we had pinned mimalloc 🤦