是尾递归吗？ [英] is it tail recursion
问题描述
int harmonic（int n）{
if（n = 1）{
返回1;
}
else {
返回谐波（n-1）+ 1 / n;
}
}
可以帮助我吗？
是Tail Rercursion吗？
i只想得到关于尾递归的答案和一些例子
尾部递归，这样我就可以区分尾部和非b / b
尾部累积。
int harmonic(int n) {
if (n=1) {
return 1;
}
else {
return harmonic(n-1)+1/n;
}
}
can any help me ??
Is it Tail Rercursion??
i just want the answer about tail recursion and and some examples
of tail recursion so that i can differentiate between tail and non-
tail recrsion.
推荐答案
您好，
我相信，根据定义，它不是。在第二种情况下，
函数不会返回_exactly_递归
调用返回的值（有一个必须添加和除法操作
在返回值之前执行
。
有关定义，请参阅：
http://www.google.com/search?hl=en&q...tail+recursion
问候。
Hello,
I believe that, by definition, it is not. In the second case the
function does not return _exactly_ the value returned by a recursive
call (there is an addition and a division operation that have to be
performed before returning the value).
For definitions, see:
http://www.google.com/search?hl=en&q...tail+recursion
Regards.
On 2008-11-03 03:55:06 -0500，Muzammil< mu ** ***********@gmail.comsaid：
On 2008-11-03 03:55:06 -0500, Muzammil <mu*************@gmail.comsaid:
int harmonic（int n）{
if（n = 1）{
返回1;
}
else {
返回谐波（n-1）+ 1 / n;
}
}
可以帮助我吗？
是吗Tail Rercursion ??
i只想得到关于尾递归的答案和一些例子
的尾递归让我可以区分尾巴和非尾巴之间的转换。
int harmonic(int n) {
if (n=1) {
return 1;
}
else {
return harmonic(n-1)+1/n;
}
}
can any help me ??
Is it Tail Rercursion??
i just want the answer about tail recursion and and some examples
of tail recursion so that i can differentiate between tail and non-
tail recrsion.
如果稍微更改最后一行，答案会更清晰：
返回1 / n +谐波（n -1）;
由于函数的尾部是对函数的调用，这是尾部
递归。它可以被一个简单的循环取代，这是许多编译器所做的优化。
-
Pete
Roundhouse Consulting，Ltd。（ www.versatilecoding.com ）作者
标准C ++库扩展：教程和参考
（ www.petebecker.com/tr1book ）
If you change the last line slightly the answer becomes a bit clearer:
return 1/n + harmonic(n-1);
Since the tail of the function is a call to the function, this is tail
recursion. It could be replaced by a simple loop, which is an
optimization that many compilers do.
--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)
On 2008-11-03 06:18:44 -0500， al**********@gmail.com 说：
On 2008-11-03 06:18:44 -0500, al**********@gmail.com said:
您好，
我相信，根据定义，它不是。在第二种情况下，
函数不会返回_exactly_递归
调用返回的值（有一个必须添加和除法操作
在返回值之前执行
。
有关定义，请参阅：
http://www.google.com/search?hl=en&q...tail+recursion
问候。
Hello,
I believe that, by definition, it is not. In the second case the
function does not return _exactly_ the value returned by a recursive
call (there is an addition and a division operation that have to be
performed before returning the value).
For definitions, see:
http://www.google.com/search?hl=en&q...tail+recursion
Regards.
大多数编译器应该能够通过识别尾递归来优化原始代码到
循环。这不是返回
的问题，而是递归调用返回的值（这将是一个非常小的函数集，无用），但是是否可以将递归调用
更改为循环。一个例子不能是Ackerman的
函数：
int ackerman（unsigned i，unsigned j）
{
if（i == 0）
返回j + 1;
否则if（j == 0）
返回ackerman（i - 1,1）;
else
返回ackerman（i - 1，ackerman（i，j - 1））;
}
结果取决于多个递归调用，因此函数不能将
转换为一个简单的循环。它也快速增长;观看
溢出。
-
Pete
Roundhouse Consulting，Ltd。（ www.versatilecoding.com ）
标准的作者C ++库扩展：教程和参考
（ www.petebecker.com / tr1book ）
Most compilers ought to be able to optimize the original code into a
loop by recognizing the tail recursion. It''s not a matter of returning
exactly the value returned by the recursive call (that would be a very
small set of functions, all useless), but of whether the recursive call
can be changed into a loop. An example that can''t is Ackerman''s
function:
int ackerman(unsigned i, unsigned j)
{
if (i == 0)
return j + 1;
else if (j == 0)
return ackerman(i - 1, 1);
else
return ackerman(i - 1, ackerman(i, j - 1));
}
The result depends on multiple recursive calls, so the function can''t
be transformed into a simple loop. It also grows very quickly; watch
out for overflows.
--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)
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