C# 针对多种类型检查对象类型

Question

有没有办法将类型数组传递给“is”运算符？

我正在尝试简化针对多种类型检查对象的语法。

就像是：

public static function bool IsOfType(object Obj,params Type[] Types)

然而，这将需要以下用法：

if(X.IsOfType(typeof(int),typeof(float))
{...}

我想做类似的事情：

if(X is {int,float})

或者

if(X.IsOfType(int,float))

甚至

public static bool ISOfType<T[]>(this object Obj){...}
if(X.ISOfType<int,float>())

我认为它们都是不可能的。

Answer 1

使用 C# 10，您可以使用模式匹配来测试对象的多种类型。

例子：

public bool TestObject<T>(T obj)
{
    if (obj is not null and (int or float or uint))
    {
        // obj is not null and guaranteed to be an int, float or uint 
    }
}

示例 2（如果您想使用该值）：

public bool TestIntOrFloat<T>(T obj)
{
    if (obj is not (int or float))
    {
        // obj is neither int or float
    }
    else if (obj is int i)
    {
        // obj is an integer and can be used with the variable 'i'
    }
    else if (obj is float f)
    {
        // obj is a floating number and can be used with the variable 'f'
    }
}

Answer 2

如果您可以将类型作为通用参数传递，那么有一个解决方案。 不幸的是，C# 不支持可变参数泛型。 您必须为每个通用参数定义函数。

public static bool IsOfType<T>(this object obj) => obj is T;
public static bool IsOfType<T1, T2>(this object obj) => obj is T1 || obj is T2;
public static bool IsOfType<T1, T2, T3>(this object obj) => obj is T1 || obj is T2 || obj is T3;
public static bool IsOfType<T1, T2, T3, T4>(this object obj) => obj is T1 || obj is T2 || obj is T3 || obj is T4;
public static bool IsOfType<T1, T2, T3, T4, T5>(this object obj) => obj is T1 || obj is T2 || obj is T3 || obj is T4 || obj is T5;
public static bool IsOfType<T1, T2, T3, T4, T5, T6>(this object obj) => obj is T1 || obj is T2 || obj is T3 || obj is T4 || obj is T5 || obj is T6;
public static bool IsOfType<T1, T2, T3, T4, T5, T6, T7>(this object obj) => obj is T1 || obj is T2 || obj is T3 || obj is T4 || obj is T5 || obj is T6 || obj is T7;
public static bool IsOfType<T1, T2, T3, T4, T5, T6, T7, T8>(this object obj) => obj is T1 || obj is T2 || obj is T3 || obj is T4 || obj is T5 || obj is T6 || obj is T7 || obj is T8;

我怀疑你需要超过 8 种类型，但如果你这样做，只需定义更多的重载。

Answer 3

它看起来很疯狂，但您可以使用以下 Fluent 语法：

object someInstance = 5.0;
if(someInstance
    .Is<int>()
    .Or<double>()) {
    // ...
}

这里 fluent-syntax 实现如下：

static class FluentIs {
    public static IsResult Is<T>(this object target) {
        return new IsResult(target, () => target is T);
    }
    public static IsResult Or<T>(this IsResult prev) {
        return new IsResult(prev, (v, x) => v || (x is T));
    }
    public class IsResult {
        Func<bool> value;
        object target;
        internal IsResult(IsResult r, Func<bool, object, bool> getValue) :
            this(r.target, () => getValue(r.value(), r.target)) {
        }
        internal IsResult(object target, Func<bool> value) {
            this.target = target;
            this.value = value;
        }
        // bool Operators
        public static bool operator true(IsResult r) { return r.value(); }
        public static bool operator false(IsResult r) { return !r.value(); }
    }
}

Answer 4

我的“选择”解决方案是使用我方便的“In”扩展功能根据对象类型名称进行验证。

    public static bool IsAnyOf(this object Obj,params string[] TypeNames)
    {
        return Obj.GetType().Name.In(TypeNames);
    }
    public static bool In(this string Needle,params string [] Haystack)
    {
        foreach (string straw in Haystack)
            if (straw == Needle)
                return true;
        return false;
    }

用法：

    public static void CheckAll(this TreeNode Node)
    {
        foreach(TreeNode Child in Node.Nodes)
        {
            if(Child.GetType().Name.In("ChildTablesNode","ChildTableNode"))
            {
                Child.Checked = Node.Checked;
                Child.CheckAll();
            }
        }
    }

    private void ctxTree_Opening(object sender, CancelEventArgs e)
    {
        TreeNode nSelected = Tree.SelectedNode;

        ctxScript.Visible = !nSelected.IsAnyOf("TablesNode"
            , "ViewsNode"
            , "ProceduresNode"
            , "UserTableTypesNode"
            , "FunctionsNode"
            , "ServerNode"
            , "DatabaseNode"
            , "ColumnsNode"
            , "ColumnNode"
            , "TriggersNode");

        Type[] x = { typeof(int) };

    }