替换lambda表达式中的参数

Question

考虑这段代码:

public class Foo
{
    public int a { get; set; }
    public int b { get; set; }
}

private void Test()
{
    List<Foo> foos = new List<Foo>();
    foos.Add(new Foo());
    foos.Add(new Foo());
    Expression<Func<Foo, int>> exp0 = f => f.a * f.b;
    Expression<Func<int>> exp1 = () => foos[0].a * foos[0].b;
    Expression<Func<int>> exp2 = () => foos[1].a * foos[1].b;
}

你怎么能把exp0它变成两个与exp1和相同的表达式exp2.请注意,我不想仅为exp0每个Fooin 评估foos,而是获得两个新表达式.

[更新]:

基本上,我希望能够扩大或"扁平化"传递给A的表达Linq扩展方法,如Sum入枚举每个项目一个表达,因为这些枚举是静态的,因为我已经有代码读取不表达获取参数(然后将它们转换为另一种语言).

我使用它MetadataToken作为对具有特定属性的属性的引用(在这种情况下a并b具有此属性)并将其与将C#属性与另一种语言的变量相关联的字典一起使用:

Foo foo = new Foo();
Expression<Func<int>> exp = () => foo.a * foo.a + foo.b;
string result1 = GetResult(exp); // gets "v_001 * v_001 + v_002"

List<Foo> foes = new List<Foo>();
foes.Add(new Foo());
foes.Add(new Foo());
Expression<Func<int>> exp2 = () => foes.Sum(f => f.a * f.a + f.b);
string result2 = GetResult(exp2); // should get "(v_001 * v_001 + v_002) + (v_003 * v_003 + v_004)"

Answer 1

我会这样做:

编写一个参数 - replacer表达式 - 访问者,操作原始表达式如下:

1. 从lambda签名中删除完全不需要的参数.
2. 用所需的索引器表达式替换参数的所有用法.

这是一个快速而肮脏的样本,我根据我之前对另一个问题的回答掀起了一个问题:

public static class ParameterReplacer
{
    // Produces an expression identical to 'expression'
    // except with 'source' parameter replaced with 'target' expression.     
    public static Expression<TOutput> Replace<TInput, TOutput>
                    (Expression<TInput> expression,
                    ParameterExpression source,
                    Expression target)
    {
        return new ParameterReplacerVisitor<TOutput>(source, target)
                    .VisitAndConvert(expression);
    }

    private class ParameterReplacerVisitor<TOutput> : ExpressionVisitor
    {
        private ParameterExpression _source;
        private Expression _target;

        public ParameterReplacerVisitor
                (ParameterExpression source, Expression target)
        {
            _source = source;
            _target = target;
        }

        internal Expression<TOutput> VisitAndConvert<T>(Expression<T> root)
        {
            return (Expression<TOutput>)VisitLambda(root);
        }

        protected override Expression VisitLambda<T>(Expression<T> node)
        {
            // Leave all parameters alone except the one we want to replace.
            var parameters = node.Parameters
                                 .Where(p => p != _source);

            return Expression.Lambda<TOutput>(Visit(node.Body), parameters);
        }

        protected override Expression VisitParameter(ParameterExpression node)
        {
            // Replace the source with the target, visit other params as usual.
            return node == _source ? _target : base.VisitParameter(node);
        }
    }
}

您的方案的用法(已测试):

var zeroIndexIndexer = Expression.MakeIndex
        (Expression.Constant(foos),
         typeof(List<Foo>).GetProperty("Item"), 
         new[] { Expression.Constant(0) });


// .ToString() of the below looks like the following: 
//  () =>    (value(System.Collections.Generic.List`1[App.Foo]).Item[0].a
//         *  value(System.Collections.Generic.List`1[App.Foo]).Item[0].b)
var exp1Clone = ParameterReplacer.Replace<Func<Foo, int>, Func<int>>
                  (exp0, exp0.Parameters.Single(), zeroIndexIndexer);