如何写一个'过布尔值for`环（false和true） [英] How to write a `for` loop over bool values (false and true)

问题描述

一个大多为乐趣/好奇的问题：如何编写C ++中的为循环会遍历的两个值布尔（即真正和假），只用行动与 BOOL （即没有转换成其它类型的）？

A question mostly for fun/curiosity: how to write a for loop in C++ that would iterate over two values of a bool (i.e. true and false), using only operations with bool (i.e. without conversions to other types)?

的背景是，我想检查有多少解决方案的公式存在像（A和和B）|| （B＆安培;！＆安培;！C＆放大器;＆安培;！D）==真，并开始写类似为（布尔A = FALSE; ???; ++ A）为（布尔B = FALSE; ......）等，但立即得到了由卡住??? - 即会产生什么的状态继续循环？当然，我重写它使用int和我也知道，做...而循环的工作，但我得到了好奇，如果这是有史以来可能写出这样的为循环？而且，由于SO似乎并不有一个答案，我决定问：）

The background is that I wanted to check how many solutions exists for an equation like (A && B) || (!B && !C && !D) == true, and started to write something like for (bool A=false; ??? ; ++A) for (bool B=false; ...) etc but immediately got stuck by ??? - i.e. what would be the condition to continue the loop? Of course I rewrote it to use int, and I also know that a do ... while loop will work, but I got curious if it's ever possible to write such a for loop? And since SO does not seem to have an answer, I decided to ask :)

更新：注意，显而易见的变种的（！布尔A = FALSE; A; A = TRUE）中至少有两个现在已经删除的的答案建议只会运行一个迭代，因为第二个条件！A 变成假和循环结束。

Update: note that an "obvious" variant for(bool A=false; !A; A=true) suggested in at least two now-removed answers will only run one iteration, because for the second one the condition !A becomes false and the loop ends.

经过一番琢磨之后，我相信这是不可能做到这一点在C ++ 03没有迪特马尔库尔提出了第二个变量或基于指针的结构等。该条件应在所需执行测试三次，这样一个布尔值的两个值是不够的。而do-whil​​e循环工作，因为第一次迭代无条件执行，条件只有两次检查等一个布尔值可用于之间持续和退出。

After some pondering, I believe it's impossible to do it in C++03 without a second variable or a pointer based construct like suggested by Dietmar Kühl. The condition should be tested three times in a desired execution, so two values of a bool are simply not enough. And the do-while loop works because the first iteration is executed unconditionally, the condition is only checked twice and so a bool value can be used to select between continuing and exiting.

推荐答案

在C ++ 11：为（布尔乙：{假，真}）{/ * ... * /}

结果

下面是一个C ++ 03版：

Here's a C++03 version:

for (bool a = true, b = false; b != a; a = a && b, b = !b) { /*...*/ }

（或者使用 A 或 B ）

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