有人可以告诉我哪里出错了？ [英] Can someone please tell me where I am going wrong??
问题描述
我正在尝试编写一个程序，要求用户在区间[1,2]中输入一个数字
，程序然后给出
$ b $的自然对数b那个数字，使用log（x + 1）系列......
这是我到目前为止所做的并且无法弄清楚我做错了什么。
任何帮助将不胜感激，谢谢你们...
#include< stdio.h>
＃包括< math.h>
int main（）{
int i，n;
double x;
浮点数，期限;
printf（"在区间[1,2]中输入一个数字）;
scanf（"％d"，& x）;
sum = 0;
i = 0;
做{
i = i + 1;
x =（double）i;
term = pow（x，i ）/ i;
if（i％2 == 1）
sum = sum + term;
else
sum = sum - term;
} while（i< = n）;
printf（&the; wer is％lf \ n"，sum）;
E-Dot写道：
我正在尝试编写一个程序，要求用户在区间[1,2]中输入一个数字
，程序然后给出
这个数字，使用日志系列（x + 1）......
这是我到目前为止所不知道的是什么我我做错了。
任何帮助都会非常感谢，谢谢你们......
#include< stdio.h>
#include< math.h>
int main（）{
如果主要没有去要接受命令行参数，请指定void。
int i，n;
double x;
float总和，期限;
为什么不将sum和term加倍？
printf（"输入一个数字区间[1,2]）;
使用换行符结束printf字符串或在printf之后调用fflush（stdout）
。否则，您的输出可能会延迟，因为
的缓冲。
scanf（"％d"，& x）;
你告诉scanf寻找一个十进制整数值并将它存储到一个双重对象中。你确定这是你想要的吗？
sum = 0;
i = 0;
做{
i = i + 1;
x =（double）i;
你在x中覆盖你以前的值。
term = pow（x， I）/ I;
由于您之前的任务，x和i将具有相同的值。
if（i ％2 == 1）
sum = sum + term;
else
sum = sum - term;
} while（i< = n）;
您将i与未初始化的值进行比较，（n）;未定义
行为。
printf（"答案是％lf \ n"，sum）;
对于打印浮点对象，请使用％f格式说明符。 ％lf
格式说明符添加了C99，作为替代，但许多
编译器还没有完全支持它。
E-Dot写道：
我正在尝试编写一个程序，要求用户输入一个数字
在区间[1,2]中，程序然后给出该数字的自然对数
，使用log（x + 1）系列...
如果你发布了所有作业会更好。
这就是我所拥有的远远不能弄清楚我做错了什么。
也许你说的是症状是什么......出了什么问题？
#include< stdio.h>
#include< math.h>
int main（）{
int i，n;
double x;
浮动金额，期限;
printf（"在区间[1,2]"）中输入一个数字;
scanf（"％d"，& x）;
sum = 0;
i = 0;
做{
i = i + 1;
可以使用（结果相同，但输入更少）
i ++;
x =（double）i;
term = pow（x，i）/ i;
你可以避免在循环中调用pow（）。在pow（x，i）和pow（x，i + 1）之间有什么区别
，是否有模式？
if（i％2 == 1）
sum = sum + term;
或
sum + = term;
else
sum = sum - term;
} while（i< = n）;
这不会给你六位小数吗？
并且n没有初始化。
< blockquote class =post_quotes>
>
printf（"答案是％lf \ n"，sum）;
如果你做了作业所说的并打印了x，sum和log（x）
你可能已经能够自己调试这个（x会错的）
-
Nick Keighley
Nick Keighley写道：
E-Dot写道：
我正在写一个程序要求用户在区间[1,2]中输入一个数字
，程序然后给出该数字的自然对数
，使用该系列进行日志（x + 1）...
请注意''s log（x + 1），而不是log（x）...
< snip>
-
Nick Keighley
I am trying to write a program which asks the user to enter a number
in the interval [1,2], the program then gives the natural logarithm of
that number, using the series for log(x+1)...
Here is what I have so far and can''t figure out what i''m doing wrong.
any help would be greatly appreciated, thanks guys...
#include <stdio.h>
#include <math.h>
int main() {
int i,n;
double x;
float sum, term;
printf("Enter a number in the interval [1,2]");
scanf("%d", &x);
sum = 0;
i = 0;
do {
i = i+1;
x = (double) i;
term = pow(x, i)/i;
if(i % 2 == 1)
sum = sum + term;
else
sum = sum - term;
} while(i <= n);
printf("The answer is %lf\n",sum);
E-Dot wrote:I am trying to write a program which asks the user to enter a number
in the interval [1,2], the program then gives the natural logarithm of
that number, using the series for log(x+1)...
Here is what I have so far and can''t figure out what i''m doing wrong.
any help would be greatly appreciated, thanks guys...
#include <stdio.h>
#include <math.h>
int main() {If main is not going to accept command line parameters, specify void.
int i,n;
double x;
float sum, term;Why not make sum and term double as well?
printf("Enter a number in the interval [1,2]");Either end the printf string with a newline or call fflush(stdout)
just after printf. Otherwise, your output may appear delayed because
of buffering.
scanf("%d", &x);You''re telling scanf to look for a decimal integer value and store it
into a double object. Are you sure that this is what you want?
sum = 0;
i = 0;
do {
i = i+1;
x = (double) i;And you''re overwriting your previous value in x.
term = pow(x, i)/i;Here x and i will have the same value due to your previous assignment.
if(i % 2 == 1)
sum = sum + term;
else
sum = sum - term;
} while(i <= n);You''re comparing i against an uninitialised value, (n); Undefined
behaviour.
printf("The answer is %lf\n",sum);For printing a float-point object use the %f format specifier. The %lf
format specifier was added with C99, as an alternative, but many
compilers don''t yet fully support it.
E-Dot wrote:
I am trying to write a program which asks the user to enter a number
in the interval [1,2], the program then gives the natural logarithm of
that number, using the series for log(x+1)...it would be better if you posted all of the assignment.
Here is what I have so far and can''t figure out what i''m doing wrong.perhaps if you said what the symptom was... What is going wrong?
#include <stdio.h>
#include <math.h>
int main() {
int i,n;
double x;
float sum, term;
printf("Enter a number in the interval [1,2]");
scanf("%d", &x);
sum = 0;
i = 0;
do {
i = i+1;could use (same result, but less to type)
i++;
x = (double) i;
term = pow(x, i)/i;you can avoid calling pow() in the loop. What is the difference
between pow(x,i) and pow(x,i+1), is there a pattern?
if(i % 2 == 1)
sum = sum + term;or
sum += term;
else
sum = sum - term;
} while(i <= n);this isn''t going to give you six decimal places is it?
And n is not initialised.
>
printf("The answer is %lf\n",sum);if you did what the assignment said and printed x, sum and log(x)
you might have been able to debug this yourself (x would be wrong)
--
Nick Keighley
Nick Keighley wrote:E-Dot wrote:
I am trying to write a program which asks the user to enter a number
in the interval [1,2], the program then gives the natural logarithm of
that number, using the series for log(x+1)...
note that''s log(x + 1), not log(x)...
<snip>
--
Nick Keighley
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