[英]sqrt function in C prints -nan
While doing some code excercise, I observed unusual ouput caused by the sqrt funtion, 在执行一些代码练习时,我观察到由sqrt函数引起的异常输出,
The code was, 代码是
#include<stdio.h>
#include<math.h>
int main()
{
double l,b,min_r,max_r;
int i;
scanf("%lf %lf",&b,&l);
printf("%lf %lf\n",sqrt(l*l+b*b),sqrt(b*b-l*l));
return(0);
}
Output: 输出:
4 5
6.403124 -nan
Why does this happenes. 为什么会这样。
Look at the numbers: b
is 4 and l
is 5. So b*b - l*l
is -9. 看一下数字:
b
是4, l
是5。所以b*b - l*l
是-9。 What's the square root of -9? -9的平方根是多少? It's an imaginary number, but
sqrt
doesn't support imaginary results, so the result is nan (not a number). 这是一个虚数,但
sqrt
不支持虚数结果,因此结果为nan(不是数字)。 It's a domain error. 这是域错误。 The solution: Don't pass negative arguments to
sqrt
. 解决方案:不要将否定参数传递给
sqrt
。
In your case, not validating the inputs cause the issue. 在您的情况下,不验证输入会导致问题。
sqrt(b*b-l*l)
with b
as 4 and l
as 5 produces a -ve number, which is most possibly you don't want. 将
b
设为4并将l
设为5会产生-ve数,这很可能是您不想要的。
FWIW, root of a negative number needs imaginary part to be represented. FWIW,负数的根需要虚部表示。
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