为什么一些C#λ表达式会编译成静态方法？
正如您在下面的代码中看到的，我已经将一个Action<>对象声明为一个变量。
谁能让我知道为什么这个action方法委托的行为像一个静态方法？
为什么它在下面的代码中返回true？
代码：
public static void Main(string[] args)
{
Action<string> actionMethod = s => { Console.WriteLine("My Name is " + s); };
Console.WriteLine(actionMethod.Method.IsStatic);
Console.Read();
}输出：
回答 4
Stack Overflow用户
发布于 2014-09-01 18:43:25
这很可能是因为没有闭包，例如：
int age = 25;
Action<string> withClosure = s => Console.WriteLine("My name is {0} and I am {1} years old", s, age);
Action<string> withoutClosure = s => Console.WriteLine("My name is {0}", s);
Console.WriteLine(withClosure.Method.IsStatic);
Console.WriteLine(withoutClosure.Method.IsStatic);这将输出withClosure的false和withoutClosure的true。
当你使用lambda表达式时，编译器会创建一个小类来包含你的方法，这将编译成如下所示(实际的实现很可能略有不同)：
private class <Main>b__0
{
public int age;
public void withClosure(string s)
{
Console.WriteLine("My name is {0} and I am {1} years old", s, age)
}
}
private static class <Main>b__1
{
public static void withoutClosure(string s)
{
Console.WriteLine("My name is {0}", s)
}
}
public static void Main()
{
var b__0 = new <Main>b__0();
b__0.age = 25;
Action<string> withClosure = b__0.withClosure;
Action<string> withoutClosure = <Main>b__1.withoutClosure;
Console.WriteLine(withClosure.Method.IsStatic);
Console.WriteLine(withoutClosure.Method.IsStatic);
}您可以看到生成的Action<string>实例实际上指向这些生成的类上的方法。
Stack Overflow用户
发布于 2015-07-27 19:09:40
在Roslyn中更改了委派缓存行为。以前，如上所述，任何没有捕获变量的lambda表达式都会在调用点被编译成static方法。Roslyn改变了这种行为。现在，任何捕获变量或不捕获变量的lambda都被转换为一个显示类：
在这个例子中：
public class C
{
public void M()
{
var x = 5;
Action<int> action = y => Console.WriteLine(y);
}
}本机编译器输出：
public class C
{
[CompilerGenerated]
private static Action<int> CS$<>9__CachedAnonymousMethodDelegate1;
public void M()
{
if (C.CS$<>9__CachedAnonymousMethodDelegate1 == null)
{
C.CS$<>9__CachedAnonymousMethodDelegate1 = new Action<int>(C.<M>b__0);
}
Action<int> arg_1D_0 = C.CS$<>9__CachedAnonymousMethodDelegate1;
}
[CompilerGenerated]
private static void <M>b__0(int y)
{
Console.WriteLine(y);
}
}罗斯林：
public class C
{
[CompilerGenerated]
private sealed class <>c__DisplayClass0
{
public static readonly C.<>c__DisplayClass0 CS$<>9__inst;
public static Action<int> CS$<>9__CachedAnonymousMethodDelegate2;
static <>c__DisplayClass0()
{
// Note: this type is marked as 'beforefieldinit'.
C.<>c__DisplayClass0.CS$<>9__inst = new C.<>c__DisplayClass0();
}
internal void <M>b__1(int y)
{
Console.WriteLine(y);
}
}
public void M()
{
Action<int> arg_22_0;
if (arg_22_0 = C.
<>c__DisplayClass0.CS$<>9__CachedAnonymousMethodDelegate2 == null)
{
C.<>c__DisplayClass0.CS$<>9__CachedAnonymousMethodDelegate2 =
new Action<int>(C.<>c__DisplayClass0.CS$<>9__inst.<M>b__1);
}
}
}Delegate caching behavior changes in Roslyn谈到了为什么要做出这样的改变。
Stack Overflow用户
发布于 2017-04-05 05:01:10
从C# 6开始，这将始终默认为实例方法，并且永远不会是静态的(因此actionMethod.Method.IsStatic将始终为false)。
查看此处：Why has a lambda with no capture changed from a static in C# 5 to an instance method in C# 6?
这里：Difference in CSC and Roslyn compiler's static lambda expression evaluation?
https://stackoverflow.com/questions/25603965
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