在 try { return x; 中真正发生了什么?} 最后 { x = n

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;
}

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