我看到这个提示的另外一个问题,想知道是否有人能向我解释,地球这个工程是如何呢?
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,在最后
子句真正执行的 的的返回
语句之后?如何线程安全的是这样的code?你能想到的任何额外的两轮牛车,可以做WRT这个尝试,终于
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?
没有 - 在IL水平,你不能从一个异常处理块内返回。它实质上是将其存储在一个变量,并返回之后
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):
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
}
这基本上声明一个局部变量( CS $ 1 $ 0000
),则以值到变量(处理块内),然后退出模块负载变量后,然后返回它。反射呈现此为:
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;
}