问题
Consider the following code:
// module level declaration
Socket _client;
void ProcessSocket() {
_client = GetSocketFromSomewhere();
using (_client) {
DoStuff(); // receive and send data
Close();
}
}
void Close() {
_client.Close();
_client = null;
}
Given that that the code calls the Close() method, which closes the _client socket and sets it to null, while still inside the `using' block, what exactly happens behind the scenes? Does the socket really get closed? Are there side effects?
P.S. This is using C# 3.0 on the .NET MicroFramework, but I suppose the c#, the language, should function identically. The reason i am asking is that occasionally, very rarely, I run out of sockets (which is a very precious resource on a .NET MF devices).
回答1:
Dispose will still be called. All you are doing is pointing the variable _client to something else in memory (in this case: null). The object that _client intially referred to will still be disposed at the end of the using statement.
Run this example.
class Program
{
static Foo foo = null;
static void Main(string[] args)
{
foo = new Foo();
using (foo)
{
SomeAction();
}
Console.Read();
}
static void SomeAction()
{
foo = null;
}
}
class Foo : IDisposable
{
#region IDisposable Members
public void Dispose()
{
Console.WriteLine("disposing...");
}
#endregion
}
Setting the variable to null is not destroying the object or preventing it from being disposed by the using. All you are doing is changing the reference of the variable, not changing the object originally referenced.
Late edit:
Regarding a discussion from the comments about MSDN's using reference http://msdn.microsoft.com/en-us/library/yh598w02.aspx and the code in the OP and in my example, I created a simpler version of the code like this.
Foo foo = new Foo();
using (foo)
{
foo = null;
}
(And, yes, the object still gets disposed.)
You could infer from the link above that the code is being rewritten like this:
Foo foo = new Foo();
{
try
{
foo = null;
}
finally
{
if (foo != null)
((IDisposable)foo).Dispose();
}
}
Which would not dispose the object, and that does not match the behavior of the code snippet. So I took a look at it through ildasm, and the best I can gather is that the original reference is being copied into a new address in memory. The statement foo = null; applies to the original variable, but the call to .Dispose() is happening on the copied address. So here is a look at how I believe the code is actually being rewritten.
Foo foo = new Foo();
{
Foo copyOfFoo = foo;
try
{
foo = null;
}
finally
{
if (copyOfFoo != null)
((IDisposable)copyOfFoo).Dispose();
}
}
For reference, this is what the IL looks like through ildasm.
.method private hidebysig static void Main() cil managed
{
.entrypoint
// Code size 29 (0x1d)
.maxstack 1
.locals init ([0] class Foo foo,
[1] class Foo CS$3$0000)
IL_0000: newobj instance void Foo::.ctor()
IL_0005: stloc.0
IL_0006: ldloc.0
IL_0007: stloc.1
.try
{
IL_0008: ldnull
IL_0009: stloc.0
IL_000a: leave.s IL_0016
} // end .try
finally
{
IL_000c: ldloc.1
IL_000d: brfalse.s IL_0015
IL_000f: ldloc.1
IL_0010: callvirt instance void [mscorlib]System.IDisposable::Dispose()
IL_0015: endfinally
} // end handler
IL_0016: call int32 [mscorlib]System.Console::Read()
IL_001b: pop
IL_001c: ret
} // end of method Program::Main
I don't make a living staring at ildasm, so my analysis can be classified as caveat emptor. However, the behavior is what it is.
回答2:
I suppose you could figure this out by looking at the disassembly, but it's a lot easier to just read section 8.13 of the specification, where all these rules are clearly described.
Reading those rules makes it clear that the code
_client = GetSocketFromSomewhere();
using (_client)
{
DoStuff();
Close();
}
is transformed by the compiler into
_client = GetSocketFromSomewhere();
{
Socket temp = _client;
try
{
DoStuff();
Close();
}
finally
{
if (temp != null) ((IDispose)temp).Dispose();
}
}
So that's what happens. The socket gets disposed twice in the non-exceptional code path. This strikes me as probably not fatal, but definitely a bad smell. I'd write this as:
_client = GetSocketFromSomewhere();
try
{
DoStuff();
}
finally
{
Close();
}
It's perfectly clear that way and nothing gets double-closed.
回答3:
As Anthony pointed out Dispose() will be called even if the reference is nulled during the execution of the using block. If you take a look at the generated IL, you'll see that even tough ProcessSocket() uses an instance member to store the field, a local reference is still created on the stack. It is via this local reference that Dispose() is called.
The IL for ProcessSocket() looks like this
.method public hidebysig instance void ProcessSocket() cil managed
{
.maxstack 2
.locals init (
[0] class TestBench.Socket CS$3$0000)
L_0000: ldarg.0
L_0001: ldarg.0
L_0002: call instance class TestBench.Socket TestBench.SocketThingy::GetSocketFromSomewhere()
L_0007: stfld class TestBench.Socket TestBench.SocketThingy::_client
L_000c: ldarg.0
L_000d: ldfld class TestBench.Socket TestBench.SocketThingy::_client
L_0012: stloc.0
L_0013: ldarg.0
L_0014: call instance void TestBench.SocketThingy::DoStuff()
L_0019: ldarg.0
L_001a: call instance void TestBench.SocketThingy::Close()
L_001f: leave.s L_002b
L_0021: ldloc.0
L_0022: brfalse.s L_002a
L_0024: ldloc.0
L_0025: callvirt instance void [mscorlib]System.IDisposable::Dispose()
L_002a: endfinally
L_002b: ret
.try L_0013 to L_0021 finally handler L_0021 to L_002b
}
Notice the local and notice how this is set to point to the member on lines L_000d-L_0012. The local is loaded again in L_0024 and used to call Dispose() in L_0025.
回答4:
using just translates to a simple try/finally where in the finally block _client.Dispose() is called if _client isn't null.
so since you close _client and set it to null, the using doesn't really do anything when it closes.
来源：https://stackoverflow.com/questions/2513035/dynamics-of-the-using-keyword