知道为什么这个程序没有产生输出？ [英] Any idea why this program is not generating an output ?
问题描述
#include< stdio.h>
#include< math.h>
#define M_PI 3.14159
int main（无效）
{
双theta，phi，sinth;
重复计算;
double incr;
双倍;
s = 180 / M_PI; / *转换为弧度* /
incr = 0.5;
theta = 0;
for（theta = incr; theta < 180; theta + = incr）
sinth = sin（s * theta）;
for（phi = 0; phi <360; phi + = incr / sinth）
count ++;
printf（"％f"，count）;
返回0;
}
#include <stdio.h>
#include <math.h>
#define M_PI 3.14159
int main(void)
{
double theta, p sinth;
double count;
double incr;
double s;
s = 180/M_PI; /* converting to radians */
incr = 0.5;
theta = 0;
for(theta = incr; theta < 180; theta += incr)
sinth = sin(s *theta);
for(phi = 0; phi < 360 ; phi += incr/ sinth)
count ++;
printf("%f", count);
return 0;
}
推荐答案
lector写道：
[主题：知道为什么这个程序没有生成输出？]
lector wrote:
[subject: Any idea why this program is not generating an output ?]
#include< stdio.h>
#include< math.h>
#define M_PI 3.14159
int main（无效）
{
双theta，phi，sinth;
重复计算;
#include <stdio.h>
#include <math.h>
#define M_PI 3.14159
int main(void)
{
double theta, p sinth;
double count;
lector.c：8：警告：''count''可以在这个
函数中未初始化使用
lector.c:8: warning: ''count'' may be used uninitialized in this
function
double incr;
double s;
s = 180 / M_PI; / *转换为弧度* /
double incr;
double s;
s = 180/M_PI; /* converting to radians */
如果你_sivide_ by s，也许。
If you _divide_ by s, perhaps.
incr = 0.5 ;
theta = 0;
for（theta = incr; theta< 180; theta + = incr）
sinth = sin（s * theta）;
incr = 0.5;
theta = 0;
for(theta = incr; theta < 180; theta += incr)
sinth = sin(s *theta);
尝试添加...
printf（"％f\ n"，sinth）;
再次运行，然后问问自己为什么不打印。
Try adding...
printf("%f\n", sinth);
Run it again, then ask yourself why it doesn''t print.
for（phi = 0; phi <360 ; phi + = incr / sinth）
count ++;
printf（"％f"，count）;
返回0;
}
for(phi = 0; phi < 360 ; phi += incr/ sinth)
count ++;
printf("%f", count);
return 0;
}
-
Peter
--
Peter
4月8日，06：53，lector< hannibal.lecto ... @ gmail.comwrote：
On 8 Apr, 06:53, lector <hannibal.lecto...@gmail.comwrote:
#include< stdio.h>
#include< math.h>
#define M_PI 3.14159
#include <stdio.h>
#include <math.h>
#define M_PI 3.14159
为什么重新定义M_PI？在我的math.h中，给它一个值为3.14159265358979323846的值
。你通过重新加工它会失去很多精确度，而且不会获得任何收益。我很惊讶
你没有得到编译器警告。
Why redefine M_PI? In my math.h, it is given a value
of 3.14159265358979323846. You lose a lot of precision
by redifining it, and don''t gain anything. I''m surprised
you don''t get a compiler warning.
4月8日上午11:02，Peter Nilsson< ; ai ... @ acay.com.auwrote：
On Apr 8, 11:02 am, Peter Nilsson <ai...@acay.com.auwrote:
lector写道：
[主题：任何想法为什么会这样程序没有生成输出？]
lector wrote:
[subject: Any idea why this program is not generating an output ?]
#include< stdio.h>
#include< math.h>
#define M_PI 3.14159
#include <stdio.h>
#include <math.h>
#define M_PI 3.14159
int main（void）
{
double theta，phi，sinth;
重复计算;
int main(void)
{
double theta, p sinth;
double count;
lector.c：8：警告：''count''可以在这个
函数中未初始化使用
lector.c:8: warning: ''count'' may be used uninitialized in this
function
double incr;
double s;
double incr;
double s;
s = 180 / M_PI; / *转换为弧度* /
s = 180/M_PI; /* converting to radians */
如果你_divide_ by s，也许。
If you _divide_ by s, perhaps.
incr = 0.5 ;
theta = 0;
incr = 0.5;
theta = 0;
for（theta = incr; theta< 180; theta + = incr）
sinth = sin（s * THETA）;
for(theta = incr; theta < 180; theta += incr)
sinth = sin(s *theta);
尝试添加...
printf（"％f\ n"，sinth）;
再次运行，然后问问自己为什么不打印。
Try adding...
printf("%f\n", sinth);
Run it again, then ask yourself why it doesn''t print.
for（phi = 0; phi <360 ; phi + = incr / sinth）
count ++;
printf（"％f"，count）;
返回0;
}
for(phi = 0; phi < 360 ; phi += incr/ sinth)
count ++;
printf("%f", count);
return 0;
}
*修改*
#include< stdio。 h>
#include< math.h>
#define M_PI 3.14159
int main（无效）
{
双theta，phi，sinth;
重复计算;
double incr;
double s;
s = 180 / M_PI;
incr = 0.05;
theta = 0;
count = 0;
for（theta = incr; theta< 180; theta + = incr）
{sinth = sin（theta / s）; printf（"％f\ n，sinth）;
for（phi = 0; phi< 360; phi + = incr / sinth）
count + +;
}
printf（"％f"，count）;
返回0;
}
运行得非常好。谢谢。
*modified*
#include <stdio.h>
#include <math.h>
#define M_PI 3.14159
int main(void)
{
double theta, p sinth ;
double count;
double incr;
double s;
s = 180/M_PI;
incr = 0.05;
theta = 0;
count = 0;
for(theta = incr; theta < 180; theta += incr)
{ sinth = sin(theta/s); printf("%f\n",sinth);
for(phi = 0; phi < 360 ; phi += incr/ sinth)
count ++;
}
printf("%f", count);
return 0;
}
Runs perfectly well. Thanks.
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