我想做的是:
GetString(
inputString,
ref Client.WorkPhone)
private void GetString(string inValue, ref string outValue)
{
if (!string.IsNullOrEmpty(inValue))
{
outValue = inValue;
}
}
这给了我一个编译错误。我想我想达到的目的已经很清楚了。基本上我想要GetString复制输入字符串的内容到客户端的WorkPhone属性。
是否可以通过引用传递属性?
当前回答
我用ExpressionTree变体和c#7写了一个包装器(如果有人感兴趣的话):
public class Accessor<T>
{
private Action<T> Setter;
private Func<T> Getter;
public Accessor(Expression<Func<T>> expr)
{
var memberExpression = (MemberExpression)expr.Body;
var instanceExpression = memberExpression.Expression;
var parameter = Expression.Parameter(typeof(T));
if (memberExpression.Member is PropertyInfo propertyInfo)
{
Setter = Expression.Lambda<Action<T>>(Expression.Call(instanceExpression, propertyInfo.GetSetMethod(), parameter), parameter).Compile();
Getter = Expression.Lambda<Func<T>>(Expression.Call(instanceExpression, propertyInfo.GetGetMethod())).Compile();
}
else if (memberExpression.Member is FieldInfo fieldInfo)
{
Setter = Expression.Lambda<Action<T>>(Expression.Assign(memberExpression, parameter), parameter).Compile();
Getter = Expression.Lambda<Func<T>>(Expression.Field(instanceExpression,fieldInfo)).Compile();
}
}
public void Set(T value) => Setter(value);
public T Get() => Getter();
}
像这样使用它:
var accessor = new Accessor<string>(() => myClient.WorkPhone);
accessor.Set("12345");
Assert.Equal(accessor.Get(), "12345");
其他回答
是的，你不能传递一个属性，但你可以将你的属性转换为带有支持字段的属性，并做类似的事情。
public class SomeClass
{
private List<int> _myList;
public List<int> MyList
{
get => return _myList;
set => _myList = value;
}
public ref List<int> GetMyListByRef()
{
return ref _myList;
}
}
但有更好的解决方案，比如动作委托等。
如果你想同时获取和设置属性，你可以在c# 7中使用这个:
GetString(
inputString,
(() => client.WorkPhone, x => client.WorkPhone = x))
void GetString(string inValue, (Func<string> get, Action<string> set) outValue)
{
if (!string.IsNullOrEmpty(outValue.get()))
{
outValue.set(inValue);
}
}
没有复制属性
void Main()
{
var client = new Client();
NullSafeSet("test", s => client.Name = s);
Debug.Assert(person.Name == "test");
NullSafeSet("", s => client.Name = s);
Debug.Assert(person.Name == "test");
NullSafeSet(null, s => client.Name = s);
Debug.Assert(person.Name == "test");
}
void NullSafeSet(string value, Action<string> setter)
{
if (!string.IsNullOrEmpty(value))
{
setter(value);
}
}
我用ExpressionTree变体和c#7写了一个包装器(如果有人感兴趣的话):
public class Accessor<T>
{
private Action<T> Setter;
private Func<T> Getter;
public Accessor(Expression<Func<T>> expr)
{
var memberExpression = (MemberExpression)expr.Body;
var instanceExpression = memberExpression.Expression;
var parameter = Expression.Parameter(typeof(T));
if (memberExpression.Member is PropertyInfo propertyInfo)
{
Setter = Expression.Lambda<Action<T>>(Expression.Call(instanceExpression, propertyInfo.GetSetMethod(), parameter), parameter).Compile();
Getter = Expression.Lambda<Func<T>>(Expression.Call(instanceExpression, propertyInfo.GetGetMethod())).Compile();
}
else if (memberExpression.Member is FieldInfo fieldInfo)
{
Setter = Expression.Lambda<Action<T>>(Expression.Assign(memberExpression, parameter), parameter).Compile();
Getter = Expression.Lambda<Func<T>>(Expression.Field(instanceExpression,fieldInfo)).Compile();
}
}
public void Set(T value) => Setter(value);
public T Get() => Getter();
}
像这样使用它:
var accessor = new Accessor<string>(() => myClient.WorkPhone);
accessor.Set("12345");
Assert.Equal(accessor.Get(), "12345");
受Sven的表达式树解决方案的启发，下面是一个不依赖反射的简化版本。此外，它还删除了不必要的自定义getter和字段表达式。
using System;
using System.Linq.Expressions;
namespace Utils;
public class Accessor<T>
{
public Accessor(Expression<Func<T>> expression)
{
if (expression.Body is not MemberExpression memberExpression)
throw new ArgumentException("expression must return a field or property");
var parameterExpression = Expression.Parameter(typeof(T));
_setter = Expression.Lambda<Action<T>>(Expression.Assign(memberExpression, parameterExpression), parameterExpression).Compile();
_getter = expression.Compile();
}
public void Set(T value) => _setter(value);
public T Get() => _getter();
private readonly Action<T> _setter;
private readonly Func<T> _getter;
}
