将方法组转换为表达式

发布于 2024-07-24 07:24:41 字数 866 浏览 6 评论 0原文

我试图弄清楚是否有一个简单的语法可以将方法组转换为表达式。使用 lambda 似乎很容易，但它不能转换为方法：

鉴于

public delegate int FuncIntInt(int x);

以下所有内容都是有效的：

Func<int, int> func1 = x => x;
FuncIntInt del1 = x => x;
Expression<Func<int, int>> funcExpr1 = x => x;
Expression<FuncIntInt> delExpr1 = x => x;

但如果我对实例方法尝试相同的操作，它会在表达式处崩溃：

Foo foo = new Foo();
Func<int, int> func2 = foo.AFuncIntInt;
FuncIntInt del2 = foo.AFuncIntInt;
Expression<Func<int, int>> funcExpr2 = foo.AFuncIntInt; // does not compile
Expression<FuncIntInt> delExpr2 = foo.AFuncIntInt;      //does not compile

最后两个都无法编译“无法将方法组 'AFuncIntInt' 转换为非委托类型 'System.Linq.Expressions.Expression<...>'。您打算调用该方法吗？”

那么有没有一个好的语法来捕获表达式中的方法组呢？

谢谢，阿恩

原文

I'm trying to figure out of if there is a simple syntax for converting a Method Group to an expression. It seems easy enough with lambdas, but it doesn't translate to methods:

Given

public delegate int FuncIntInt(int x);

all of the below are valid:

Func<int, int> func1 = x => x;
FuncIntInt del1 = x => x;
Expression<Func<int, int>> funcExpr1 = x => x;
Expression<FuncIntInt> delExpr1 = x => x;

But if i try the same with an instance method, it breaks down at the Expressions:

Foo foo = new Foo();
Func<int, int> func2 = foo.AFuncIntInt;
FuncIntInt del2 = foo.AFuncIntInt;
Expression<Func<int, int>> funcExpr2 = foo.AFuncIntInt; // does not compile
Expression<FuncIntInt> delExpr2 = foo.AFuncIntInt;      //does not compile

Both of the last two fail to compile with "Cannot convert method group 'AFuncIntInt' to non-delegate type 'System.Linq.Expressions.Expression<...>'. Did you intend to invoke the method?"

So is there a good syntax for capturing a method grou in an expression?

thanks,
arne

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千笙结 2024-07-31 07:24:42

我使用属性而不是方法。

public class MathLibrary
{
    public Expression<Func<int, int>> AddOne {  
        get {   return input => input + 1;} 
    }
}

使用上面

I use property instead of method.

public class MathLibrary
{
    public Expression<Func<int, int>> AddOne {  
        get {   return input => input + 1;} 
    }
}

Using above

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时光倒影 2024-07-31 07:24:41

这个怎么样？

  Expression<Func<int, int>> funcExpr2 = (pArg) => foo.AFuncIntInt(pArg);
  Expression<FuncIntInt> delExpr2 = (pArg) => foo.AFuncIntInt(pArg);

How about this?

  Expression<Func<int, int>> funcExpr2 = (pArg) => foo.AFuncIntInt(pArg);
  Expression<FuncIntInt> delExpr2 = (pArg) => foo.AFuncIntInt(pArg);

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鱼忆七猫命九 2024-07-31 07:24:41

也可以使用 NJection.LambdaConverter 委托到 LambdaExpression 转换器库来完成此操作

public class Program
{
    private static void Main(string[] args) {
       var lambda = Lambda.TransformMethodTo<Func<string, int>>()
                          .From(() => Parse)
                          .ToLambda();            
    }   

    public static int Parse(string value) {
       return int.Parse(value)
    } 
}

It is also possible to do it using NJection.LambdaConverter a Delegate to LambdaExpression converter Library

public class Program
{
    private static void Main(string[] args) {
       var lambda = Lambda.TransformMethodTo<Func<string, int>>()
                          .From(() => Parse)
                          .ToLambda();            
    }   

    public static int Parse(string value) {
       return int.Parse(value)
    } 
}

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