属性不能通过引用传递。这里有一些方法可以绕过这个限制。

1. 返回值

string GetString(string input, string output)
{
    if (!string.IsNullOrEmpty(input))
    {
        return input;
    }
    return output;
}

void Main()
{
    var person = new Person();
    person.Name = GetString("test", person.Name);
    Debug.Assert(person.Name == "test");
}

2. 委托

void GetString(string input, Action<string> setOutput)
{
    if (!string.IsNullOrEmpty(input))
    {
        setOutput(input);
    }
}

void Main()
{
    var person = new Person();
    GetString("test", value => person.Name = value);
    Debug.Assert(person.Name == "test");
}

3.LINQ表达式

void GetString<T>(string input, T target, Expression<Func<T, string>> outExpr)
{
    if (!string.IsNullOrEmpty(input))
    {
        var expr = (MemberExpression) outExpr.Body;
        var prop = (PropertyInfo) expr.Member;
        prop.SetValue(target, input, null);
    }
}

void Main()
{
    var person = new Person();
    GetString("test", person, x => x.Name);
    Debug.Assert(person.Name == "test");
}

4. 反射

void GetString(string input, object target, string propertyName)
{
    if (!string.IsNullOrEmpty(input))
    {
        var prop = target.GetType().GetProperty(propertyName);
        prop.SetValue(target, input);
    }
}

void Main()
{
    var person = new Person();
    GetString("test", person, nameof(Person.Name));
    Debug.Assert(person.Name == "test");
}

为了对这个问题进行投票，这里有一个关于如何将其添加到语言中的积极建议。我并不是说这是最好的方法，请随意提出你自己的建议。但是允许像Visual Basic那样通过ref传递属性将极大地帮助简化一些代码，而且经常如此!

https://github.com/dotnet/csharplang/issues/1235

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);
    }
}

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");

这在c#语言规范的7.4.1节中介绍。只有变量引用可以作为参数列表中的ref或out形参传递。属性不符合变量引用的条件，因此不能使用。

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;
}