问题描述
我在另一个问题中看到了这个提示,想知道是否有人可以向我解释这到底是如何工作的?
I saw this tip in another question and was wondering if someone could explain to me how on earth this works?
try { return x; } finally { x = null; }
我的意思是,finally 子句真的执行 after return 语句吗?这段代码有多线程不安全?你能想到任何可以做的额外黑客行为吗?这个 try-finally 破解?
I mean, does the finally clause really execute after the return statement? How thread-unsafe is this code? Can you think of any additional hackery that can be done w.r.t. this try-finally hack?
推荐答案
不 - 在 IL 级别,您不能从异常处理块内部返回.它本质上将它存储在一个变量中,然后返回
No - at the IL level you can't return from inside an exception-handled block. It essentially stores it in a variable and returns afterwards
即类似于:
int tmp; try { tmp = ... } finally { ... } return tmp;
例如(使用反射器):
static int Test() { try { return SomeNumber(); } finally { Foo(); } }
编译为:
.method private hidebysig static int32 Test() cil managed { .maxstack 1 .locals init ( [0] int32 CS$1$0000) L_0000: call int32 Program::SomeNumber() L_0005: stloc.0 L_0006: leave.s L_000e L_0008: call void Program::Foo() L_000d: endfinally L_000e: ldloc.0 L_000f: ret .try L_0000 to L_0008 finally handler L_0008 to L_000e }
这基本上声明了一个局部变量(CS$1$0000),将值放入变量中(在处理的块内),然后在退出块后加载变量,然后返回它.反射器将其呈现为:
This basically declares a local variable (CS$1$0000), places the value into the variable (inside the handled block), then after exiting the block loads the variable, then returns it. Reflector renders this as:
private static int Test() { int CS$1$0000; try { CS$1$0000 = SomeNumber(); } finally { Foo(); } return CS$1$0000; }