c#枚举函数参数
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c# Enum Function Parameters
提问by TK.
As a follow on from this question.
How can I call a function and pass in an Enum?
如何调用函数并传入枚举？
For example I have the following code:
例如，我有以下代码：
enum e1
{
//...
}
public void test()
{
myFunc( e1 );
}
public void myFunc( Enum e )
{
var names = Enum.GetNames(e.GetType());
foreach (var name in names)
{
// do something!
}
}
Although when I do this I am getting the 'e1' is a 'type' but is used like a 'variable' Error message. Any ideas to help?
虽然当我这样做时，我得到的 'e1' 是一个 'type'，但它被用作一个 'variable' 错误消息。有什么想法可以帮助吗？
I am trying to keep the function generic to work on any Enum not just a specific type? Is this even possible?... How about using a generic function? would this work?
我试图保持函数通用以适用于任何 Enum 而不仅仅是特定类型？这甚至可能吗？...如何使用通用函数？这行得通吗？
采纳答案by bruno conde
You can use a generic function:
您可以使用通用函数：
public void myFunc<T>()
{
var names = Enum.GetNames(typeof(T));
foreach (var name in names)
{
// do something!
}
}
and call like:
并调用：
myFunc<e1>();
(EDIT)
（编辑）
The compiler complains if you try to constraint Tto Enumor enum.
如果您尝试约束T到Enum或，编译器会抱怨enum。
So, to ensure type safety, you can change your function to:
因此，为了确保类型安全，您可以将函数更改为：
public static void myFunc<T>()
{
Type t = typeof(T);
if (!t.IsEnum)
throw new InvalidOperationException("Type is not Enum");
var names = Enum.GetNames(t);
foreach (var name in names)
{
// do something!
}
}
回答by Andrew Hare
You are trying to pass the type of the enum as an instance of that type - try something like this:
您正在尝试将枚举类型作为该类型的实例传递 - 尝试如下操作：
enum e1
{
foo, bar
}
public void test()
{
myFunc(e1.foo); // this needs to be e1.foo or e1.bar - not e1 itself
}
public void myFunc(Enum e)
{
foreach (string item in Enum.GetNames(e.GetType()))
{
// Print values
}
}
回答by Martin Moser
Why not passing the type? like:
为什么不传递类型？喜欢：
myfunc(typeof(e1));
public void myFunc( Type t )
{
}
回答by abatishchev
Use
用
public void myFunc( e1 e ) { // use enum of type e}
instead of
代替
public void myFunc( Enum e ) { // use type enum. The same as class or interface. This is not generic! }
