Question

我有一个函数foo，它采用非托管类型，然后我创建了一个通用结构，它需要类型参数不受管理：

[<Struct>]
type Vector4<'T when 'T:unmanaged> =
    val x : 'T
    val y : 'T
    val z : 'T
    val w : 'T
    new (x, y, z, w) = { x = x; y = y; z = z; w = w }

let foo<'T when 'T:unmanaged> (o:'T) =
    printfn "%A" o
    printfn "%d" sizeof<'T>

let bar() =
    let o = Vector4<float32>(1.0f, 2.0f, 3.0f, 4.0f)
    foo o  // here has error

但我得到了编译错误：

Error 4 A generic construct requires that the type 'Vector4<float32>' is an unmanaged type

我检查了MSDN，它是syas：

提供的类型必须是非托管类型。非托管类型某些原始类型（sbyte，byte，char，nativeint， unativeint，float32，float，int16，uint16，int32，uint32，int64， uint64或十进制），枚举类型，nativeptr＆lt; _＆gt;或非泛型结构，其字段都是非托管类型。

为什么需要blittable类型参数的泛型结构不是非托管类型？

Answer 1

Interop不支持通用类型：[1]，[2]

COM模型不支持泛型类型的概念。因此，泛型类型不能直接用于COM互操作。

不幸的是，在这种情况下，类型别名没有帮助：

[<Struct>]
[<StructLayout(LayoutKind.Sequential)>]
type Vector4<'T when 'T:unmanaged> =
    val x : 'T
    val y : 'T
    val z : 'T
    val w : 'T
    new (x, y, z, w) = { x = x; y = y; z = z; w = w }

type Vector4float = Vector4<float32>

let inline foo<'T when 'T:unmanaged> (o:'T) =
    printfn "%A" o
    printfn "%d" sizeof<'T>

let bar() =
    let o = new Vector4float(1.0f, 2.0f, 3.0f, 4.0f)
    foo o //  A generic construct requires that the type 'Vector4float' is an unmanaged type

为什么由blittable类型参数化的泛型结构不是非托管类型？

1 个答案: