Java处理stackoverflow，并在stackoverflow错误后继续正常执行

Question

I'm trying to do a recursion in Java. I just want to stop the recursion and continue normal prgram execution

void doit(){
    try{
        doit();
    }
    catch (StackOverflowError e) {
        return;
    }
    System.out.println("Error");
}


statement1
doit()
statementcontinue

I want the program to continue execution to statementcontinue after the stackoverflow error

Answer 1

Your program is doing exactly what you told it to.

Each time you call doit() , it:

1. Calls doit() again
2. After that finishes, it prints Error .

When the stack overflow happens, the innermost call finishes (because of your return ), and then continues executing the function that called it (like any other function call).
This is called popping the call stack.

The calling function (which is also doit() ) then executes the next line ( System.out.println("Error"); ), then returns to its calling function, which is also doit() .
The cycle repeats until the stack is fully popped – until it gets up tothe function that originally called doit() .

Answer 2

If you want to print "Error" only when the stackOverflow occurs, just place the trace in the catch block:

void doit(){
  try{
    doit();
  }catch (StackOverflowError e) {
    System.out.println("Error");
    return;
  }
}

Answer 3

Your code fills up the stack, then once the stack is full it hits the catch statement. After that the rest of the code continues to fire... each of those error messages is a recursive call that was made. Your code is working as you programmed it.

If you want an example of recursion that does stuff before and after, and has an exit condition, the following should work as an example for you (with print statements to clarify what is happening on the stack).

Example:

public class RecurseExample {

    public static void main(String[] args) {
        System.out.println("hi");
        doIt(1);
        System.out.println("bye");
    }

    private static void doIt(int i){
        if (i <= 3){
            System.out.println("i before: " + i);
            doIt(++i);
            System.out.println("i after: " + i);
        } else {
            System.out.println("this is where the recursion stops and the execution will continue to unravel the stack");
        }
    }
}

The output:

hi
i before: 1
i before: 2
i before: 3
this is where the recursion stops and the execution will continue to unravel the stack
i after: 4
i after: 3
i after: 2
bye