生成参数化F＃报价

Question

Let's say we have a simple F# quotation:

type Pet = {  Name : string }
let exprNonGeneric = <@@ System.Func(fun (x : Pet) -> x.Name) @@>

The resulting quotation is like:

val exprNonGeneri : Expr =
  NewDelegate (System.Func`2[[FSI_0152+Pet, FSI-ASSEMBLY, Version=0.0.0.0, Culture=neutral, PublicKeyToken=null],[System.String, mscorlib, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089]],
             x, PropertyGet (Some (x), System.String Name, []))

Now I want to generalize it, so I instead of type "Pet" and property "Name" I could use an arbitrary type and method/property defined on it. Here is what I am trying to do:

let exprGeneric<'T, 'R> f = <@@ System.Func<'T, 'R>( %f ) @@>
let exprSpecialized = exprGeneric<Pet, string> <@ (fun (x : Pet) -> x.Name) @>

The resulting expression is now different:

val exprSpecialized : Expr =
  NewDelegate (System.Func`2[[FSI_0152+Pet, FSI-ASSEMBLY, Version=0.0.0.0, Culture=neutral, PublicKeyToken=null],[System.String, mscorlib, Version=4.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089]],
             delegateArg,
             Application (Lambda (x,
                                  PropertyGet (Some (x), System.String Name, [])),
                          delegateArg))

As you can see, the difference between the first and the second expression is that in first case the top level NewDelegate expression contains PropertyGet, while the second expression wraps PropertyGet in a Application/Lambda expression. And when I pass this expression to an external code it does not expect such expression structure and fails.

So I need some way to build a generalized version of quotation, so when it gets specialized, the resulting quotation is an exact match of <@@ System.Func(fun (x : Pet) -> x.Name) @@>. Is this possible? Or it there only choice to manually apply pattern matching to a generated quotation and transform it to what I need?

UPDATE . As a workaround I implemented the following adapter:

let convertExpr (expr : Expr) =
    match expr with
    | NewDelegate(t, darg, appl) ->
        match (darg, appl) with
        | (delegateArg, appl) ->
            match appl with 
            | Application(l, ldarg) ->
                match (l, ldarg) with
                | (Lambda(x, f), delegateArg) ->
                    Expr.NewDelegate(t, [x], f)
                | _ -> expr
            | _ -> expr
    | _ -> expr

It does the job - I can now convert expression from 1st to 2nd form. But I am interested in finding out if this can be achieved in a simple way, without traversing expression trees.

Answer 1

I don't think it will be possible to do this; in the second case, you are plugging in the expression <@ (fun (x : Pet) -> x.Name) @> , which is represented using a Lambda node, into the hole in the other expression. The compiler does not simplify expressions during this plugging process, so the Lambda node won't be removed no matter what you do.

However your pattern matching workaround can be greatly simplified:

let convertExpr = function
| NewDelegate(t, [darg], Application(Lambda(x,f), Var(arg))) 
    when darg = arg -> Expr.NewDelegate(t, [x], f)
| expr -> expr

In fact, your more complicated version is incorrect. This is because the delegateArg in your innermost pattern is not matching against the value of the previously bound delegateArg identifier from the outer pattern; it is a new, freshly bound identifier which also happens to be called delegateArg . In fact, the outer delegateArg identifier has type Var list while the inner one has type Expr ! However, given the limited range of expression forms generated by the compiler your broken version may not be problematic in practice.

EDIT

Regarding your followup questions, if I understand you correctly it may not be possible to achieve what you want. Unlike C#, where x => x + 1 could be interpreted as having a type of either Func<int,int> or Expression<Func<int,int>> , in F# fun x -> x + 1 is always of type int->int . If you want to get a value of type Expr<int->int> then you generally need to use the quotation operator (<@ @>) .

There is one alternative that may be of use, however. You can use the [<ReflectedDefinition>] attribute on let bound functions to make their quotations available as well. Here's an example:

open Microsoft.FSharp.Quotations
open Microsoft.FSharp.Quotations.ExprShape
open Microsoft.FSharp.Quotations.Patterns
open Microsoft.FSharp.Quotations.DerivedPatterns

let rec exprMap (|P|_|) = function
| P(e) -> e
| ShapeVar(v) -> Expr.Var v
| ShapeLambda(v,e) -> Expr.Lambda(v, exprMap (|P|_|) e)
| ShapeCombination(o,l) -> RebuildShapeCombination(o, l |> List.map (exprMap (|P|_|)))


let replaceDefn = function
| Call(None,MethodWithReflectedDefinition(e),args) 
    -> Some(Expr.Applications(e, [args]))
| _ -> None


(* plugs all definitions into an expression *)
let plugDefs e = exprMap replaceDefn e

[<ReflectedDefinition>]
let f x = x + 1

(* inlines f into the quotation since it uses the [<ReflectedDefinition>] attribute *)
let example = plugDefs <@ fun y z -> (f y) - (f 2) @>