下面的C＃是一个非常简单的循环，但我认为它是两个循环。我的一位同事说，他认为这是一个单一的循环。你能告诉我是一个循环还是两个循环？您还能告诉我如何阅读IL并向我的同事证明这是两个循环吗？
var ints = new List<int> {1, 2, 3, 4};
foreach (var i in ints.Where(x => x != 2))
{
Console.WriteLine(i);
}
如果事实证明这实际上是一个很酷的循环。我仍然想知道如何读取IL并看到它只是一个循环。
.method private hidebysig static void Main(string[] args) cil managed
{
.entrypoint
// Code size 137 (0x89)
.maxstack 3
.locals init ([0] class [mscorlib]System.Collections.Generic.List`1<int32> ints,
[1] int32 i,
[2] class [mscorlib]System.Collections.Generic.List`1<int32> '<>g__initLocal0',
[3] class [mscorlib]System.Collections.Generic.IEnumerator`1<int32> CS$5$0000,
[4] bool CS$4$0001)
IL_0000: nop
IL_0001: newobj instance void class [mscorlib]System.Collections.Generic.List`1<int32>::.ctor()
IL_0006: stloc.2
IL_0007: ldloc.2
IL_0008: ldc.i4.1
IL_0009: callvirt instance void class [mscorlib]System.Collections.Generic.List`1<int32>::Add(!0)
IL_000e: nop
IL_000f: ldloc.2
IL_0010: ldc.i4.2
IL_0011: callvirt instance void class [mscorlib]System.Collections.Generic.List`1<int32>::Add(!0)
IL_0016: nop
IL_0017: ldloc.2
IL_0018: ldc.i4.3
IL_0019: callvirt instance void class [mscorlib]System.Collections.Generic.List`1<int32>::Add(!0)
IL_001e: nop
IL_001f: ldloc.2
IL_0020: ldc.i4.4
IL_0021: callvirt instance void class [mscorlib]System.Collections.Generic.List`1<int32>::Add(!0)
IL_0026: nop
IL_0027: ldloc.2
IL_0028: stloc.0
IL_0029: nop
IL_002a: ldloc.0
IL_002b: ldsfld class [mscorlib]System.Func`2<int32,bool> ConsoleApplication1.Program::'CS$<>9__CachedAnonymousMethodDelegate2'
IL_0030: brtrue.s IL_0045
IL_0032: ldnull
IL_0033: ldftn bool ConsoleApplication1.Program::'<Main>b__1'(int32)
IL_0039: newobj instance void class [mscorlib]System.Func`2<int32,bool>::.ctor(object,
native int)
IL_003e: stsfld class [mscorlib]System.Func`2<int32,bool> ConsoleApplication1.Program::'CS$<>9__CachedAnonymousMethodDelegate2'
IL_0043: br.s IL_0045
IL_0045: ldsfld class [mscorlib]System.Func`2<int32,bool> ConsoleApplication1.Program::'CS$<>9__CachedAnonymousMethodDelegate2'
IL_004a: call class [mscorlib]System.Collections.Generic.IEnumerable`1<!!0> [System.Core]System.Linq.Enumerable::Where<int32>(class [mscorlib]System.Collections.Generic.IEnumerable`1<!!0>,
class [mscorlib]System.Func`2<!!0,bool>)
IL_004f: callvirt instance class [mscorlib]System.Collections.Generic.IEnumerator`1<!0> class [mscorlib]System.Collections.Generic.IEnumerable`1<int32>::GetEnumerator()
IL_0054: stloc.3
.try
{
IL_0055: br.s IL_0067
IL_0057: ldloc.3
IL_0058: callvirt instance !0 class [mscorlib]System.Collections.Generic.IEnumerator`1<int32>::get_Current()
IL_005d: stloc.1
IL_005e: nop
IL_005f: ldloc.1
IL_0060: call void [mscorlib]System.Console::WriteLine(int32)
IL_0065: nop
IL_0066: nop
IL_0067: ldloc.3
IL_0068: callvirt instance bool [mscorlib]System.Collections.IEnumerator::MoveNext()
IL_006d: stloc.s CS$4$0001
IL_006f: ldloc.s CS$4$0001
IL_0071: brtrue.s IL_0057
IL_0073: leave.s IL_0087
} // end .try
finally
{
IL_0075: ldloc.3
IL_0076: ldnull
IL_0077: ceq
IL_0079: stloc.s CS$4$0001
IL_007b: ldloc.s CS$4$0001
IL_007d: brtrue.s IL_0086
IL_007f: ldloc.3
IL_0080: callvirt instance void [mscorlib]System.IDisposable::Dispose()
IL_0085: nop
IL_0086: endfinally
} // end handler
IL_0087: nop
IL_0088: ret
} // end of method Program::Main
最佳答案
编译器将您的代码转换为try-finally块，首先在源上调用GetEnumerator方法（这是从Where)返回的迭代器，然后进入try块。
第一次检查：
IL_0055: br.s IL_0067
跳转到
IL_0067以在迭代器上调用MoveNext，然后将MoveNext的结果加载到局部变量中（正如古怪的名称所建议的那样（CS $ 4 $ 0001），这是编译器生成的变量）：IL_006d: stloc.s CS$4$0001
IL_006f: ldloc.s CS$4$0001
该指令检查从
MoveNext返回的结果是否为true，以及是否跳转回IL_0057IL_0071: brtrue.s IL_0057
然后执行继续，直到
MoveNext返回false为止一直执行相同的操作。所以是的，代码中有一个循环。您可以在documentation中找到有关
IL指令的更多信息。除此之外，
try块之前的代码可能看起来令人困惑，但它基本上会创建一个Func<int, bool>委托，这是您的lambda表达式（x => x != 2），然后将其传递给Where方法，并将其结果加载到3.这行中（实际上是第四个，3是索引）局部变量：IL_0054: stloc.3
您可以在参数列表中看到
IEnumerator<int>，然后循环使用该迭代器。