递归因子函数。 [英] recursive factorial function.
问题描述
考虑（factorial.cpp）：
#include< iostream>
using namespace std;
双R = 3.2; / *未使用，但R是静态的，因为它是一个全局的
变量（文件范围）* /
int F（int n）; / *原型* /
int main（）
{
int number;
cout << 输入一个整数，其中找到的阶乘：" ;;
cin>>数字;
cout<< 让我们看看什么因子（<<<<<<<"）给了我们。它给了
us： << F（数量）<< endl;
}
int F（int i）/ *自动存储变量 - >在堆栈中分配（在执行此函数时仅存在
）* /
{
static int count = 0; / *计数另一方面是静态的（明确地）
- >在整个程序运行期间存在* /
++ count; / *增量计数器* /
if（i == 0）
{
cout<< Count = <<计数<< endl<<结束;
返回1;
}
其他
返回i * F（ - i）;
}
我想知道为什么它给我输出0（计数打印后）
到屏幕）...另外：我不知道如何打印出正确的
阶乘值...
祝你好运/ Med venlig hilsen
Martin J？rgensen
-
------------------ -------------------------------------------------- -------
Martin J？rgensen的主页 - http： //www.martinjoergensen.dk
" Martin J？rgensen"写道：
考虑（factorial.cpp）：
#include< iostream>
使用命名空间std;
双R = 3.2; / *未使用，但R是静态的，因为它是一个全局变量
（文件范围）* /
int F（int n）; / *原型* /
int main（）
{/ / int int number;
cout<< 输入一个整数，以查找阶乘：" ;;
cin>>数字;
cout<< 让我们看看什么因子（<<<<<<<"）给了我们。它给了我们：
<< F（数量）<< endl;
}
int F（int i）/ *自动存储变量 - >在堆栈中分配（在执行此函数时仅存在）* /
{
static int count = 0; / *计数另一方面是静态的（明确地） - >
在整个程序运行期间存在* /
++ count; / *增量计数器* /
if（i == 0）
{
cout<< Count = <<计数<< endl<<结束;
返回1;
}
返回i * F（ - i）;
}
我想知道为什么它给我输出0（在计数打印到屏幕后）......还有：我不知道如何打印出正确的因子值。 ..
你让它变得多么复杂。不需要静态的
变量，例如count。我建议你从头开始用
介意。
Martin J？rgensen skrev：你好，
考虑（factorial.cpp）：
#include< iostream>
使用命名空间std;
double R = 3.2 ; / *未使用，但R是静态的，因为它是一个全局的变量（文件范围）* /
int F（int n）; / *原型* /
< snip>
int F（int i）/ *自动存储变量 - >在堆栈中分配（在执行此函数时仅存在）* /
{
static int count = 0; / *指望另一方面是静态的
（明确地） - >在整个程序运行期间存在* /
++ count; / *增量计数器* /
if（i == 0）
{
cout<< Count = <<计数<< endl<< endl;
返回1;
}
其他
返回i * F（ - i）;
返回i * F（i - 1）;
为什么？在第二次递归''我'将等于0，并且
声明
返回0 * F（0）;
将返回0.
}
http://www.parashift.com/c++-faq-lit...html#faq-39.15
TB @ SWEDEN
" Martin J？rgensen" <未********* @ spam.jay.net>在留言中写道
news：b9 ************ @ news.tdc.dk ...
... 。
return i * F（ - i）;
我想知道为什么它给我0作为输出（计数被打印到屏幕后）...
表达式i * F（ - i）使用并修改两个
序列点之间的i值。因此，它的行为是未定义的，并且在此程序中找到* any *其他错误之前没有任何意义
，直到此错误为
更正。
Hi,
Consider (factorial.cpp):
#include <iostream>
using namespace std;
double R=3.2; /* not used, but R is static because it is a global
variable (file scope) */
int F(int n); /* prototype */
int main()
{
int number;
cout << "Enter an integer of which to find the factorial of: ";
cin >> number;
cout << "Let''s see what factorial(" << number << ") gives us. It gives
us: " << F(number) << endl;
}
int F(int i) /* automatic storage variable -> allocated in a stack (only
exists during execution of this function) */
{
static int count = 0; /* count on the other hand is static (explicitly)
-> exists during the whole program run */
++count; /* increment counter */
if (i==0)
{
cout << "Count = " << count << endl << endl;
return 1;
}
else
return i*F(--i);
}
I''m wondering why it gives me 0 as output (after count has been printed
to the screen)... Also: I''m not sure how to print out the correct
factorial value...
Best regards / Med venlig hilsen
Martin J?rgensen
--
---------------------------------------------------------------------------
Home of Martin J?rgensen - http://www.martinjoergensen.dk
"Martin J?rgensen" writes:
Consider (factorial.cpp):
#include <iostream>
using namespace std;
double R=3.2; /* not used, but R is static because it is a global variable
(file scope) */
int F(int n); /* prototype */
int main()
{
int number;
cout << "Enter an integer of which to find the factorial of: ";
cin >> number;
cout << "Let''s see what factorial(" << number << ") gives us. It gives us:
" << F(number) << endl;
}
int F(int i) /* automatic storage variable -> allocated in a stack (only
exists during execution of this function) */
{
static int count = 0; /* count on the other hand is static (explicitly) ->
exists during the whole program run */
++count; /* increment counter */
if (i==0)
{
cout << "Count = " << count << endl << endl;
return 1;
}
else
return i*F(--i);
}
I''m wondering why it gives me 0 as output (after count has been printed to
the screen)... Also: I''m not sure how to print out the correct factorial
value...
You''re making it unnecessarily complicated. There is no need for a static
variable such as count. I suggest you start from scratch with that tidbit in
mind.
Martin J?rgensen skrev:Hi,
Consider (factorial.cpp):
#include <iostream>
using namespace std;
double R=3.2; /* not used, but R is static because it is a global
variable (file scope) */
int F(int n); /* prototype */
<snip>
int F(int i) /* automatic storage variable -> allocated in a stack (only
exists during execution of this function) */
{
static int count = 0; /* count on the other hand is static
(explicitly) -> exists during the whole program run */
++count; /* increment counter */
if (i==0)
{
cout << "Count = " << count << endl << endl;
return 1;
}
else
return i*F(--i);
return i * F(i - 1);
Why? On the second last recursion ''i'' will equal 0, and
the statement
return 0 * F(0);
will return 0.
}
http://www.parashift.com/c++-faq-lit...html#faq-39.15
--
TB @ SWEDEN
"Martin J?rgensen" <un*********@spam.jay.net> wrote in message
news:b9************@news.tdc.dk...
....
return i*F(--i); I''m wondering why it gives me 0 as output (after count has been printed to
the screen)...
The expression i*F(--i) uses and modifies the value of i between two
sequence points. Therefore its behavior is undefined, and there is no point
in trying to find *any* other bugs in this program until this error is
corrected.
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