如何转换此Func& lt; SampleExpression,IEnumerator< string& gt;,bool& gt;& gt;到Func& lt; SampleExpression,bool& gt;
这是我的课程
class SampleExpression
{
public string str;
public static bool SampleEnum(SampleExpression s, IEnumerator<string> ien = null)
{
while (ien.MoveNext())
{
if (s.str == ien.Current)
{
ien.Reset();
return true;
}
}
return false;
}
}
这是我在运行时生成表达式树的方式:
This is how i am generating my expression tree at runtime:
static void Main(string[] args)
{
ParameterExpression param1 = Expression.Parameter(typeof(SampleExpression), "token");
ParameterExpression param2 = Expression.Parameter(typeof(IEnumerator<string>), "args");
var lstConstant = "1,2,3,4,".Split(new string[] { "," },
StringSplitOptions.RemoveEmptyEntries).ToList();
var enummethod = typeof(SampleExpression).GetMethod("SampleEnum");
MethodCallExpression methodCall = Expression.Call
(
enummethod,
param1
, param2
);
var e = Expression.Lambda<Func<SampleExpression, IEnumerator<string>, bool>>(methodCall, param1, param2);
var l = e.Compile();
List<SampleExpression> lst = new List<SampleExpression>();
lst.Add(new SampleExpression { str = "1" }); // matches with lstConstant
lst.Add(new SampleExpression { str = "2" }); // matches with lstConstant
lst.Add(new SampleExpression { str = "5" });
var items = lst.Where(x => l(x, lstConstant.GetEnumerator())).ToList();
}
现在我可能会以复杂的方式完成此操作(因为我是Expression Tree的新手)-我的要求是:
Now i might i have done this in a convoluted way(cause i am novice in Expression trees) - my requirement is this:
我有一个用逗号分隔的字符串,例如"1,2,3,4,"
.我想将每个 SampleExpression
与类 SampleExpression
的字符串参数 str
进行匹配和匹配.到目前为止,我已经做到了.
I have a comma separated string like this "1,2,3,4,"
. I want to split and match each SampleExpression
with the string parameter str
of the class SampleExpression
. Which i have done so far.
但是我希望表达式为 Func< SampleExpression,bool>
.如您所见,当前它的 Func< SampleExpression,IEnumerator< string>,bool>
.
However i want the Expression as Func<SampleExpression,bool>
. As you can see currently its Func<SampleExpression, IEnumerator<string>, bool>
.
我该如何解决.
表达式编译对我来说也很奇怪,但实际上要回答您的问题...
The expression compilation seems weird to me too, but to actually answer your question...
您可以像这样包装已编译的Func:
You can wrap the compiled Func like so:
Func<SampleExpression, bool> lBind = (SampleExpression token) => l(token, lstConstant.GetEnumerator());
这会将枚举器绑定为第二个参数,同时将第一个参数保留为打开状态.
This binds the enumerator as the second parameter, while leaving the first open for your input.