haskell - 通过 GHC 的 FFI 传递一种 ByteArray 类型是什么意思？

Question

在 Haskell2010 报告中，来自 section 8.4.2 ，有一个 FFI 导入/导出声明支持的基本类型列表:

The following types constitute the set of basic foreign types:

• Char, Int, Double, Float, and Bool as exported by the Haskell Prelude as well as
• Int8, Int16, Int32, Int64, Word8, Word16, Word32, Word64, Ptr a, FunPtr a, and StablePtr a, for any type a, as exported by Foreign (Section 24).

A Haskell system that implements the FFI needs to be able to pass these types between the Haskell and the external context as function arguments and results.

来自 GHC 手册中有关 FFI 扩展的部分，GHC adds to these types with some unboxed types :

The following unboxed types may be used as basic foreign types (see FFI Addendum, Section 3.2): Int#, Word#, Char#, Float#, Double#, Addr#, StablePtr# a, MutableByteArray#, ForeignObj#, and ByteArray#.

当我指定 ccall 调用约定时，很清楚大多数这些类型在链接器级别会发生什么。例如，我很确定 Int/Int# 作为两个 32 位值在堆栈上传递。同样，Ptr a/StablePtr a/StablePtr# a/Addr# 可能都作为指针传递给堆栈。

ByteArray# 和 MutableByteArray# 呢？

我所能想象的就是将它们作为指针传递，但这似乎有点疯狂，因为除非您使 ByteArray#/MutableByteArray# 固定，否则 GHC 运行时可能最终从你下面移动阵列。此外，您还会忽略数组中的大小信息。

Answer 1

感谢@ThomasM.DuBuisson 挖掘了一个 old email thread :

There is a way to pass an unpinned ByteArray# (or MutableByteArray#, but the former seems right in your case) to a foreign call, using the UnliftedFFITypes language extension. The ByteArray# is guaranteed to not to be moved for the duration of the call. The code should treat the ByteArray# argument as if it was a pointer to bytes. You will need to do any address offset computations on the C side (i.e. pass any offsets you need as extra argument to your C function).

因此:这些作为指针传递，GHC 的运行时 promise 不会将它们从您的下方移动。