Question

是否存在任何使变量变为常量的c＃类型/结构？也许通过设置一个固定值的标志。

示例代码：

foreach (Person p in group)
{
    if (p.Mood.Equals("bad"))
    {
        group.Mood = "bad";
        group.Mood.Fix(); // This should disable any further changing on group.Mood 
    }
    else
    {
        group.Mood = "good";
    }
}

整个集团的情绪应该是＆＃34;糟糕＆＃34;如果一个人心情不好。我知道我可以自己实现这样的东西，但我经常使用它，我可以想象这样的结构已经存在。

我希望你心情愉快！ ;）

Answer 1

不，变量在声明时变为常量。但是，你可以通过属性设置器获得一些东西：

public string Mood
{
    set 
    {
       if(mood != "Bad")
            mood = value; 
    }
}

Answer 2

我建议不要这样做，原因如下：

应该允许设置带有setter的属性
带有setter的属性不应该只是忽略赋予它的值

其次，如果您已经确定了最终价值，为什么还要查看收藏中的更多人？

我会将代码更改为以下三种变体之一：

group.Mood = "good";
foreach (Person p in group)
{
    if (p.Mood.Equals("bad"))
    {
        group.Mood = "bad";
        break;
    }
}

或者这个：

string groupMood = "good";
foreach (Person p in group)
{
    if (p.Mood.Equals("bad"))
    {
        groupMood = "bad";
        break;
    }
}
group.Mood = groupMood;

或者这个：

if (group.Any(p => p.Mood == "bad"))
    group.Mood = "bad";
else
    group.Mood = "good";

如果您想要这样的“冰棒”类型，可以尝试使用LINQPad：

void Main()
{
    var i = new Popsicle<int>(10);
    i.Value = 25;
    i.Value = 33;
    i.Dump();

    i.Freeze(FrozenBehavior.IgnoreNewValues);
    i.Value = 17;
    i.Dump();

    i = new Popsicle<int>(10);
    i.Freeze(FrozenBehavior.ThrowException);
    i.Value = 17;
    i.Dump();
}

public class Popsicle<T>
{
    private T _Value;
    private FrozenBehavior? _FrozenBehavior;

    public Popsicle(T value = default(T))
    {
        _Value = value;
    }

    public T Value
    {
        get
        {
            return _Value;
        }

        set
        {
            if (_FrozenBehavior.HasValue)
            {
                switch (_FrozenBehavior.GetValueOrDefault())
                {
                    case FrozenBehavior.IgnoreNewValues:
                        break;

                    case FrozenBehavior.ThrowException:
                        throw new InvalidOperationException("This Popsicle<T> has been frozen and configured to throw exceptions if an attempt is made to change its value");

                    default:
                        throw new InvalidOperationException("This Popsicle<T> is in an invalid frozen state");
                }
            }
            else
                _Value = value;
        }
    }

    public void Freeze(FrozenBehavior behavior = FrozenBehavior.ThrowException)
    {
        if (!_FrozenBehavior.HasValue)
            _FrozenBehavior = behavior;
        else
            throw new InvalidOperationException("This Popsicle<T> has already been frozen, cannot re-freeze");
    }

    public override string ToString()
    {
        if (!_FrozenBehavior.HasValue)
            return _Value + " (unfrozen)";
        else switch (_FrozenBehavior)
        {
            case FrozenBehavior.IgnoreNewValues:
                return _Value + " (frozen: ignore)";

            case FrozenBehavior.ThrowException:
                return _Value + " (frozen: throw)";

            default:
                throw new InvalidOperationException("This Popsicle<T> is in an invalid frozen state");
        }
    }
}

public enum FrozenBehavior
{
    IgnoreNewValues,
    ThrowException
}

Answer 3

使用LinQ可以在很短的时间内完成。检查是否有任何心情不好的人，如果是，请将群组心情设置为bad，否则good。

group.Mood = group.Any(p => p.Mood == "bad") ? "bad" : "good";

将变量变为常量

3 个答案: