如何避免递归函数的StackOverflowError [英] How to avoid StackOverflowError for a recursive function

问题描述

我正在编写一个自我调用约5000次的函数。当然，我得到一个 StackOverflowError 。有什么办法可以用相当简单的方式重写这段代码吗？：

I'm writing a function that will call itself up to about 5000 times. Ofcourse, I get a StackOverflowError. Is there any way that I can rewrite this code in a fairly simple way?:

void checkBlocks(Block b, int amm) {

    //Stuff that might issue a return call

    Block blockDown = (Block) b.getRelative(BlockFace.DOWN);
    if (condition) 
        checkBlocks(blockDown, amm);


    Block blockUp = (Block) b.getRelative(BlockFace.UP);
    if (condition) 
        checkBlocks(blockUp, amm);

    //Same code 4 more times for each side

}

顺便说一句，我们可以调用函数的深度有多少限制？

By the way, what is the limitation of how deep we may call the functions?

推荐答案

使用显式堆栈对象和循环，而不是调用堆栈和递归：

Use an explicit stack of objects and a loop, rather than the call stack and recursion:

void checkBlocks(Block b, int amm) {
  Stack<Block> blocks = new Stack<Block>();
  blocks.push(b);
  while (!blocks.isEmpty()) {
    b = blocks.pop();
    Block blockDown = (Block) b.getRelative(BlockFace.DOWN);
    if (condition)
      blocks.push(block);
    Block blockUp = (Block) b.getRelative(BlockFace.UP);
    if (condition) 
      blocks.push(block);
  }
}

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