满足动态条件时退出递归函数 [英] Quit recursive function when a dynamic condition is met

问题描述

使用在汉明内生成所有比特序列的功能距离t ：

void magic(char* str, int i, int changesLeft) {
        if (changesLeft == 0) {
                printf("%s\n", str);
                return;
        }
        if (i < 0) return;
        // flip current bit
        str[i] = str[i] == '0' ? '1' : '0';
        magic(str, i-1, changesLeft-1);
        // or don't flip it (flip it again to undo)
        str[i] = str[i] == '0' ? '1' : '0';
        magic(str, i-1, changesLeft);
}

我想退出递归函数并返回调用函数发生某种情况（如果发生）。所以这就像我的递归功能是听到可能告诉她退出的声音！

I would like to quit the recursive function and return to the caller function when a certain condition occurs (if it does). So it's like my recursive function is hearing voices that might tell her to quit!

它只发生 str <打印/ code>，此处：

It only happens after str is printed, here:

if (changesLeft == 0) {
    printf("%s\n", str);
    int quit_now = voices(str);
    return;
}

如何做到这一点（停止展开递归并返回函数调用者） ？

How to do this (stop unfolding the recursion and return to the function caller)?

尝试：

if (i < 0 || quit_now == 1) return;

似乎阻止了执行并且永远不会结束！

just seems to block the execution and never end!

_{PS - 我甚至对 c感兴趣旧方法。}

_{PS - I am interested even in c old methodologies.}

推荐答案

要把它放在最简单的形式，你可以做这样的事情：

To put it in the simplest form, you could do something like this:

void foo(bool & ret) {
  // doStuff...
  if (ret) return;
  foo(ret);
  // doStuff...
  if (ret) return;
  foo(ret);
}

然后你开始递归：

bool ret = false;
foo(ret);

在您的情况下，您可以通过

In your case you can interupt the recursion by

if (!changesLeft) {
  printf("%s\n", str);
  ret = true;
  return;
}

设置为true会让你退出整个调用树。

Setting to true will get you out of the entire call tree.

您也可以在C中执行此操作，只需使用指针而不是引用。

You can do it in C as well, just use a pointer rather than a reference.

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