问题描述
我已经编写了这个用于证券分析的ob / os振荡器。唯一的
是结果总是负面的。如-96。我查看了
以及此代码，对我来说看起来很好。有没有麻烦？
#include< stdio.h>
#include< stdlib.h>
双os（双倍价格，双倍价格）
{double os;
os = price-ma / ma * 100;
返回操作系统;
}
int main（int argc，char * argv []）
{if（argc！= 3）
{fprintf（stderr，" usage error \ n"）;
退出（EXIT_FAILURE）;
}
双倍价格，马;
price = strtod（argv [1]，NULL）;
ma = strtod（argv [2]，NULL）;
printf（"％f.2 \ n"，os（price，ma））;
返回0;
}
谢谢。
I have written this ob/os oscillator for securities analysis. The only
thing is that the result is always negative. Such as -96. I have looked over
and over this code and it looks fine to me. Is there any trouble?
#include <stdio.h>
#include <stdlib.h>
double os(double price,double ma)
{ double os;
os=price-ma/ma*100;
return os;
}
int main(int argc,char *argv[])
{ if (argc!=3)
{fprintf(stderr,"usage error\n");
exit(EXIT_FAILURE);
}
double price,ma;
price=strtod(argv[1],NULL);
ma=strtod(argv[2],NULL);
printf("%f.2\n",os(price,ma));
return 0;
}
Thanks.
推荐答案
当ma是双倍时，ma / ma总是将是1（除非ma为0
或无穷大或NaN），所以你的操作程序总是会返回
价格 - 100。
*可能*你的意思是，os =（price - ma）/ ma * 100; ??
-
任何足够先进的bug都与功能无法区分。
- Rich Kulawiec
When ma is a double, ma/ma is always going to be 1 (unless ma is 0
or infinity or NaN), so your os routine is always going to return
price - 100 .
*Possibly* you meant something like, os = (price - ma) / ma * 100; ??
--
"Any sufficiently advanced bug is indistinguishable from a feature."
-- Rich Kulawiec
啊，我明白了。非常感谢沃尔特。那一定是问题所在。尚未使用
到所有C'的括号:)
Bill
Ah I see. Thanks much Walter. That must be the problem. Not quite used
to all C''s parenthesis yet :)
Bill
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