最后尝试块可以防止StackOverflowError

Question

看一下以下两种方法：

public static void foo() {
    try {
        foo();
    } finally {
        foo();
    }
}

public static void bar() {
    bar();
}

bar()清楚地运行会导致StackOverflowError，但foo()不会运行（该程序似乎无限期地运行）。 这是为什么？

Answer 1

它不会永远运行。每次堆栈溢出都会导致代码移至finally块。问题在于这将需要非常，非常长的时间。时间顺序为O（2 ^ N），其中N是最大堆栈深度。

想象最大深度为5

foo() calls
    foo() calls
       foo() calls
           foo() calls
              foo() which fails to call foo()
           finally calls
              foo() which fails to call foo()
       finally
           foo() calls
              foo() which fails to call foo()
           finally calls
              foo() which fails to call foo()
    finally calls
       foo() calls
           foo() calls
              foo() which fails to call foo()
           finally calls
              foo() which fails to call foo()
       finally
           foo() calls
              foo() which fails to call foo()
           finally calls
              foo() which fails to call foo()
finally calls
    foo() calls
       foo() calls
           foo() calls
              foo() which fails to call foo()
           finally calls
              foo() which fails to call foo()
       finally
           foo() calls
              foo() which fails to call foo()
           finally calls
              foo() which fails to call foo()
    finally calls
       foo() calls
           foo() calls
              foo() which fails to call foo()
           finally calls
              foo() which fails to call foo()
       finally
           foo() calls
              foo() which fails to call foo()
           finally calls
              foo() which fails to call foo()

要使每个级别进入finally块都需要两倍的时间，而堆栈深度可能是10,000或更大。如果您每秒可以进行10,000,000次呼叫，那么这将花费10 ^
3003秒，甚至比宇宙的时间还要长。