2

我正在尝试组合这样的功能列表。

我有这个：

Func<int, bool>[] criteria = new Func<int, bool>[3];
criteria[0] = i => i % 2 == 0;
criteria[1] = i => i % 3 == 0;
criteria[2] = i => i % 5 == 0;

我想要这个：

Func<int, bool>[] predicates = new Func<int, bool>[3];
predicates[0] = i => i % 2 == 0;
predicates[1] = i => i % 2 == 0 && i % 3 == 0;
predicates[2] = i => i % 2 == 0 && i % 3 == 0 && i % 5 == 0;

到目前为止，我有以下代码：

Expression<Func<int, bool>>[] results = new Expression<Func<int, bool>>[criteria.Length];

for (int i = 0; i < criteria.Length; i++)
{
    results[i] = f => true;
    for (int j = 0; j <= i; j++)
    {
        Expression<Func<int, bool>> expr = b => criteria[j](b);
        var invokedExpr = Expression.Invoke(
            expr, 
            results[i].Parameters.Cast<Expression>());
        results[i] = Expression.Lambda<Func<int, bool>>(
            Expression.And(results[i].Body, invokedExpr), 
            results[i].Parameters);
    }
}
var predicates = results.Select(e => e.Compile()).ToArray();

Console.WriteLine(predicates[0](6)); // Returns true
Console.WriteLine(predicates[1](6)); // Returns false
Console.WriteLine(predicates[2](6)); // Throws an IndexOutOfRangeException

有谁知道我做错了什么？

4

2 回答 2

5

无需拉入表达式...

    Func<int, bool>[] criteria = new Func<int, bool>[3];
    criteria[0] = i => i % 2 == 0;
    criteria[1] = i => i % 3 == 0;
    criteria[2] = i => i % 5 == 0;

    Func<int, bool>[] predicates = new Func<int, bool>[3];

    predicates[0] = criteria[0];
    for (int i = 1; i < criteria.Length; i++)
    {
        //need j to be an unchanging int, one for each loop execution.
        int j = i;

        predicates[j] = x => predicates[j - 1](x) && criteria[j](x);
    }

    Console.WriteLine(predicates[0](6)); //True
    Console.WriteLine(predicates[1](6)); //True
    Console.WriteLine(predicates[2](6)); //False
于 2008-10-02T12:05:54.380 回答
4

这是一个猜测，因为我对这些东西知之甚少，但这似乎可以解决它：

Func<int, bool>[] criteria = new Func<int, bool>[3]; 
criteria[0] = i => i % 2 == 0; 
criteria[1] = i => i % 3 == 0; 
criteria[2] = i => i % 5 == 0;
Expression<Func<int, bool>>[] results = new Expression<Func<int, bool>>[criteria.Length];
for (int i = 0; i < criteria.Length; i++)
{
    results[i] = f => true; 
    for (int j = 0; j <= i; j++)
    {
        int ii = i;
        int jj = j;
        Expression<Func<int, bool>> expr = b => criteria[jj](b); 
        var invokedExpr = Expression.Invoke(expr, results[ii].Parameters.Cast<Expression>()); 
        results[ii] = Expression.Lambda<Func<int, bool>>(Expression.And(results[ii].Body, invokedExpr), results[ii].Parameters);
    }
} 
var predicates = results.Select(e => e.Compile()).ToArray();

关键是'ii'和'jj'的引入（可能只有一个重要，我没试过）。我认为您在 lambda 中捕获了一个可变变量，因此当您最终引用它时，您看到的是后来变异的值而不是原始值。

于 2008-10-02T02:42:45.390 回答