LINQ的功能/表达当地评估

Question

鉴于此代码：

 int min = 0; 

     Expression<Func<List<IUser>, bool>> ulContainsJohn = 
        (l => l.Where(u => u.FirstName == "John").Count() > min); 

     Assert.AreEqual(true, ulContainsJohn.Compile()(userList)); 

     min = 3; 

     Assert.AreEqual(true, ulContainsJohn.Compile()(userList)); 

列表中包含1“约翰”，但第二断言失败。我如何将min的值急切地绑定到Func，所以它不会尝试重新评估变量min？

说明：我不想更改签名。我希望表达式树不是作为变量来评估，而是作为一个常量表达式。无论如何要转换min的评估，以便树有一个常量表达式而不是变量评估？

Answer 1

编辑：阅读你的评论，尝试函数的创造者。

Func<int, Func<List<IUser>, bool>> createFn = (min) => 
    (l) => (l.Count(u => u.FirstName == "John") > min); 

Func<List<IUser>, bool>> contains0 = createFn(0); 

Assert.AreEqual(true, contains0(userList)); 

Func<List<IUser>, bool>> contains3 = createFn(3); 

Assert.AreEqual(true, contains3(userList)); 

尝试使用1元素数组。丑，但它的作品。

var values = new int[] { 0 }; 

Expression<Func<List<IUser>, bool>> ulContainsJohn = 
       (l => l.Where(u => u.FirstName == "John").Count() > values[0]); 

Assert.AreEqual(true, ulContainsJohn.Compile()(userList)); 

values[0] = 3; 

Assert.AreEqual(true, ulContainsJohn.Compile()(userList)); 

另一种选择，更好：

private int Minimum { get; set; } 

... 

Expression<Func<List<IUser>, bool>> ulContainsJohn = 
      (l => l.Where(u => u.FirstName == "John").Count() > Minimum); 

Func<List<IUser>, bool> fn = ulContainsJohn.Compile(); 
Assert.AreEqual(true, fn(userList)); 

Minimum = 3; 

Assert.AreEqual(true, fn(userList)); 

Answer 2

最简单的解决方案是让一个独立的变量：

int min = 0; 
    int staticMin = min; 
    Func<List<IUser>, bool> ulContainsJohn = 
       l => l.Where(u => u.FirstName == "John").Count() > staticMin ; 

Answer 3

你必须使它成为一个参数。

Expression<Func<List<IUser>, int, bool>> ulContainsJohn = 
       (List<IUser> l, int min) => (l.Where(u => u.FirstName == "John").Count() > min); 


    ulContainsJohn.Compile()(userList, min);