注意，此版本甚至可以在循环自身中使用递归，两次调用循环本身，一次用于对列表左手边的内容进行排序，另一次对列表右手边的内容进行排序。
Notice that this version even makes use of recursiveness in itself, calling itself twice, once to sort the left-hand side of the list, and once again to sort the right-hand side of the list.
无论是太多的循环、递归还是其他的什么，你现在应该知道如果处理类似的情况。
Whether there be too many loops, too much recursion, or just plain too much going on, you now know how to deal with each.
因此，只有在非常复杂的大量使用循环和递归调用的应用程序中，才需要考虑这种潜在的影响。
Therefore, the potential impact is a factor only in complex applications where iterations and recursive invocations are heavily used.
在表1中了解循环的特性，看它们可以如何与递归函数的特性相对比。
Take a look at the properties of loops and see how they compare with those of recursive functions in Table 1.
一旦检测到循环，就会设置cycle子句中已定义好的适当标记，并停止分支的递归查询过程。
As soon as a cycle is detected the appropriate flag defined in the cycle clause is set and the recursive query process for the branch is stopped.
注意，在循环的结尾处，我用一个比上次大1的参数递归地调用同一个函数，以实现循环的迭代。
Note at the end of the loop that I recursively call the same function with an argument that is one greater than the previous, implementing the iteration of the loop.
通常情况是函数内包含了过多的循环(不是在循环中执行了过多的内容)，太多的递归，或者只不过是太多不相干但又要一起执行的操作。
Usually this means there's too many loops (as opposed to too much happening in a loop), too much recursion, or simply too many different operations being performed.
根据Wikipedia：递归（拉丁语recurrere->currere =run和re =return）是以自相似性的方式循环某些项的过程。
According to Wikipedia: Recursion (lat. recurrere -> currere = run and re = return) is the process of repeating items in a self-similar way.
这便是haskell没有while或for循环的原因，我们的替代方案是使用递归。
That's why there are no while loops or for loops in Haskell and instead we many times have to use recursion to declare what something is.
给出了循环几乎可分解的循环有向三元系的一些直接构造和递归构造。
The direct and recursive constructions for a cyclically almost resolvable cyclic directed triple system are given.
基于现有循环不变式开发技术中的递归定义思想，提出了开发复杂递归问题循环不变式的两种新策略；
Based on recursive definition idea in existing technique of developing loop invariants, two new strategies for developing loop invariants on complicated recursive problems are presented.
对于背包问题这样一个经典的递归问题，发现了它的不需栈支持的循环解法，并由此说明其存在并行。
For knapsack problem (which is a classical recursion problem), find its round solution which need not stack to support, then prove it exists parallel solutio.
对于背包问题这样一个经典的递归问题，发现了它的不需栈支持的循环解法，并由此说明其存在并行解。
For knapsack problem (which is a classical recursion problem), find its round solution which need not stack to support, then prove it exists parallel solution.
对于背包问题这样一个经典的递归问题，发现了它的不需栈支持的循环解法，并由此说明其存在并行解。
For knapsack problem (which is a classical recursion problem), find its round solution which need not stack to support, then prove it exists parallel solution.
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