解释所需的 C 递归

本文介绍了解释所需的 C 递归的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我需要有关此特定代码的帮助.我并没有真正完全理解它.有人可以花时间向我解释一下吗?它所做的是取一个单词并打印出它的反面.不过我不明白递归部分，对于 n=1，for 循环只会运行一次，并且递归将运行直到它读取整个单词并满足 ' ' 标记，但是它如何打印出相反的字?

I need some help with this particular code. I do not really understand it to its fullest. Could someone take the time and explain it to me? What it does is take a word and prints out its reverse. I don't understand the recursion part though, for n=1 the for loop will only run once, and the recursion will run until it reads the whole word and it meets the ' ' mark, but then how does it print out the reversed word?

void reverse() { char c; scanf("%c", &c); if (c!=' ') { reverse(); } printf("%c", c); } int main() { int n, i; printf(" The number of the words="); scanf("%d", &n); for(i=1; i<=n; ++i) { reverse(); printf(" "); } printf(" End of the program. "); return 0; }

推荐答案

NOTE 您的示例在底部，首先是您的标题问题:解释所需的 C 递归

递归是一种允许操作调用自身的编程技术.一个简单(但毫无意义)的例子是:

Recursion is a programming technique allowing operations to call themselves. A simple (but meaningless) example would be:

void call(void); int main(void) { call(); } void call(void) { call(); }

注意:这会一直持续到堆栈确定堆积了太多调用，并导致程序崩溃.

Note: This would simply go until the stack determines too many calls have stacked up, and crash the program.

一个更有意义的例子(为了说明)是写一个回文字符系列(A - Z):

A more meaningful example (for illustration) would be to write a palindrome character series (A - Z):

void Palindrome(char a) { char b[2]; sprintf(b, "%c", a); printf(b); if((a >= 65)&&(a < 90)) { Palindrome(a+1); } sprintf(b, "%c", a); printf(b); }

最后一个例子实际上做了第一个例子没有做的两件重要的事情:1) 有一个受控的退出标准if((a >= 65)&&(a <= 90))2) 将先前调用的结果与后续调用结合使用.

This last example actually does two important things the first example did not do: 1) has a controlled exit criteria if((a >= 65)&&(a <= 90)) 2) uses the results of prior calls with subsequent calls.

在你的例子中:程序工作的原因是每次操作调用自身时，它都在节中嵌套更深(每次调用一个嵌套):(对于所有递归程序都是如此)

In your example: The reason the program works is that each time the operation calls itself, it is nesting deeper (one nest for each call) into the section: (this is true for all recursive programs)

{ reverse(); }

在概念上类似于:

{ //do something { //do something { //do something //... and so on for as many recursions are necessary to meet exit criteria } } }

...但是它是如何打印出反向单词的呢?"只有在递归达到编程限制后，执行才会向下流过结束的 } 并到达下一部分，在那里它继续展开自身，每次访问时，以相反的顺序堆栈， c 的值，直到每个嵌套级别都被展开:

"...but then how does it print out the reversed word?" Only after the recursion reaches the programmed limit does execution flow down past the closing } and hit the next section, where it continues to unwind itself, each time accessing, in reverse order of stack, the values of c until each nested level has been unwound:

} //after exit criteria is met, execution flow goes from here.... printf("%c", c); //prints values of c in reverse order. } //to here until number of recursions is exhausted.

那时，如果您使用调试器逐步执行，您将看到从上层 } 到下层 } 执行 printf() 每次.

at that time, if you step through using a debugger, you will see flow going from the upper } to the lower } executing printf() each time.