try-catch 语句中的递归调用 [英] Recursive call in the try-catch statement

问题描述

我有以下代码片段:

public static void main(String[] args) {
    foo();
}
public static void foo() {
    try {
        foo();
    } catch (Throwable t) {
        foo();
    }
}

谁能解释一下这里发生了什么?请详细说明.

Can anyone explain me what's going on here? In details, please.

我已经更改了这部分并添加了 println 方法来显示某些内容:

I have changed this piece and added println methods to show something:

...
try {
    System.out.println("+");
    foo();
} catch (Throwable t) {
    System.out.println("-");
    foo();
}
...

我得到了这样的东西(这个过程还没有停止):

I'm getting something like this (the process hasn't stopped):

+
+
+
+
+--
+--
+
+--
+--
+
+
+--
+--
+
+--
+--
+
+
+
+--
+--

推荐答案

foo 退出的唯一方法是因为 stackoverflow.我们可以通过

The only way foo can exit is because of stackoverflow. We can simulate the effect by

public static void foo(int i)
{
    if(i>=N) return; // stack too deep

    foo(i+1);
    foo(i+1);
}

这对于最大堆栈深度来说是指数级的昂贵.

which is exponentially expensive w.r.t to the max stack depth.

在我的机器上，花费的时间大约是6ns * 2^N

On my machine, the time it takes is about 6ns * 2^N

最大堆栈深度超过 10,000，因此需要大约

The max stack depth is over 10,000, so it will take about

        10000000000000000000000000000000
        .... 
        (thousands of zeros)  
        ... 
        00000000000000000000000000000000 

完成该计划的年数，给予或接受一个常数因素:)

years to finish the program, give or take a constant factor:)

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