我想根据提供的字段制作一个简单的记录类型.
那是 :
let rectype = MakeRecordType(['fieldname1'; 'fieldname2'])
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直接进入类型提供者看起来像是一个如此简单的任务的重枪手.
还有其他方法吗?
更新
我发现以下问题看起来非常相似 通过反射创建F#记录
撇开最终结果的有用性,下面的一个片段实现了我在其他相关答案的精神中所要求的:
#if INTERACTIVE
#r @"C:\Program Files (x86)\Microsoft F#\v4.0\FSharp.Compiler.dll"
#r @"C:\Program Files (x86)\FSharpPowerPack-1.9.9.9\bin\FSharp.Compiler.CodeDom.dll"
#endif
open System
open System.CodeDom.Compiler
open Microsoft.FSharp.Compiler.CodeDom
open Microsoft.FSharp.Reflection
type RecordTypeMaker (typeName: string, records: (string*string) []) =
let _typeDllName = "Synth"+typeName+".dll"
let _code =
let fsCode = new System.Text.StringBuilder()
fsCode.Append("module ").Append(typeName).Append(".Code\ntype ").Append(typeName).Append(" = {") |> ignore
for rec' in records do fsCode.Append(" ").Append(fst rec').Append(" : ").Append(snd rec').Append(";\n") |> ignore
fsCode.Append("}").ToString()
let _compiled =
use provider = new FSharpCodeProvider()
let options = CompilerParameters([||], _typeDllName)
let result = provider.CompileAssemblyFromSource( options, [|_code|] )
result.Errors.Count = 0
let mutable _type: Type = null
member __.RecordType
with get() = if _compiled && _type = null then
_type <- Reflection.Assembly.LoadFrom(_typeDllName).GetType(typeName+".Code+"+typeName)
_type
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草图实现RecordTypeMaker接受Record type包含的任意定义type name和field names附带的数组field type names.然后在窗帘后面它组装了一段定义所请求记录类型的F#代码,通过编译此代码CodeDom provider,加载容器程序集并通过Reflection提供对这个新创建的合成Record类型的访问.一个测试片段
let myType = RecordTypeMaker("Test", [|("Field1", "string"); ("Field2", "int")|]).RecordType
printfn "IsRecordType=%b" (FSharpType.IsRecord(myType))
printfn "Record fields: %A" (FSharpType.GetRecordFields(myType))
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为纯粹合成类型演示了myType概念证明:
IsRecordType=true
Record fields: [|System.String Field1; Int32 Field2|]
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