代码之家 › 专栏 › 技术社区 › Ronnie Overby

c#lambda表达式+反思问题

lambda reflection .net c#

Ronnie Overby · 技术社区 · 14 年前

var mapper = new InputMapper<SomeType>();
mapper.Map("some user input", someType => someType.IntProperty, "Input was not an integer");
mapper.Map("some user input", someType => someType.BoolProperty, "Input was not a bool");

SomeType someTypeInstance = mapper.CreateInstance();

我的InputMapper类包含使用Map()方法创建的所有映射的集合。CreateInstance()将遍历试图转换用户输入的映射,并将其分配给lambda表达式中使用的属性。当它循环通过时,它将保存抛出的所有formatexception的集合。

我的问题是:

在输入映射器.Map()方法,lambda参数类型应该是什么?

谢谢!

更新

斯基特医生问我的意图。

InputMapper类将用于将用户输入分配给任何对象的成员,并负责将用户输入转换为properties类型。类的接口可以从上面的示例中推断出来。

更新2

经过一番牵手,乔恩和丹把我带到了那里。你能提出改进建议吗?以下是我所拥有的: http://pastebin.com/RaYG5n2h

3 回复 | 直到 4 年前

Jon Skeet 14 年前

对于你的第一个问题 Map

public class InputMapper<TSource> where TSource : new()
{
    public void Map<TResult>(string input,
                             Expression<Func<TSource, TResult>> projection,
                             string text)
    {
        ...
    }
}

现在值得注意的是,lambda表达式表示属性 ,但大概你想给酒店打电话 设置器 只有

一旦你 TSource 使用 new TSource() (请注意

如果没有更多的细节来说明你想做什么,恐怕很难给出更详细的答案。

var memberExpression = projection.Body as MemberExpression;
if (memberExpression == null)
{
    throw new ArgumentException("Lambda was not a member access");
}
var propertyInfo = memberExpression.Member as PropertyInfo;
if (propertyInfo == null)
{
    throw new ArgumentException("Lambda was not a property access");
}
if (projection.Parameters.Count != 1 ||
    projection.Parameters[0] != memberExpression.Expression)
{
    throw new ArgumentException("Property was not invoked on parameter");
}
if (!propertyInfo.CanWrite)
{
    throw new ArgumentException("Property is read-only");
}
// Now we've got a PropertyInfo which we should be able to write to - 
// although the setter may be private. (Add more tests for that...)
// Stash propertyInfo appropriately, and use PropertyInfo.SetValue when you 
// need to.

Ronnie Overby 14 年前

public void Map<TValue>(string input, Expression<Func<T, TValue>> property, string message);

然后CreateInstance可能看起来像:

public T CreateInstance() 
{
    T result = new T();
    foreach (var lambda in map) 
    {
        ((PropertyInfo)((MemberExpression)lambda.Body).Member).SetValue(result, null, propertyValueForLambdaThatYouStored);  
    }
    return result;
}

如果lambda中的表达式不是属性引用,则可以添加一些检查以引发更好的异常。

Dan Tao 14 年前

:参见 here

_{(

:看起来在我发布此消息之前不久,您就已经知道了。很好!不管怎样,我将把它放在那里作为参考,尽管它不是一个非常有效的实现。)}

class CustomType
{
    public int Integer { get; set; }
    public double Double { get; set; }
    public bool Boolean { get; set; }
    public string String { get; set; }

    public override string ToString()
    {
        return string.Format("int: {0}, double: {1}, bool: {2}, string: {3}", Integer, Double, Boolean, String);
    }
}

class Program
{
    public static void Main(string[] args)
    {
        var mapper = new InputMapper<CustomType>();

        mapper.Map("10", x => x.Integer, "Unable to set Integer property.");
        mapper.Map("32.5", x => x.Double, "Unabled to set Double property.");
        mapper.Map("True", x => x.Boolean, "Unable to set Boolean property.");
        mapper.Map("Hello world!", x => x.String, "Unable to set String property.");

        var customObject = mapper.Create();

        Console.WriteLine(customObject);

        Console.ReadKey();
    }
}

输出:

int: 10, double: 32.5, bool: True, string: Hello world!

听起来你想定义你的 Map

class InputMapper<T>
{
    public void Map<TProperty>(string input,
                               Expression<Func<T, TProperty>> propertyExpression,
                               string errorMessage);
}

propertyExpression

我没有清楚为什么你不想这样定义它:

class InputMapper<T>
{
    public void Map<TProperty>(string input,
                               Action<TProperty> propertySetter,
                               string errorMessage);
}

mapper.Map<int>(
    "some user input",
    value => someType.IntProperty = value,
    "Input was not an integer"
);

Map<TProperty> Convert.ChangeType .)