1.#IND as an output

[Nobody/Anonymous]

Hey everyone,
I was writing a program that solves a quadratic equation when I ran into a weird error. When entering certain values, the answer to the quadratic equation would be shown as "-1.#IND" or "1.IND". I noticed that the problem only arose when I used this line in my source:

sq = sqrt ((bb) - (4a*c));

Instead of this:

sq = sqrt (pow(b,2) - (4ac));

It seems that the "pow" function is required to make this program work. Does anyone know what "1.#IND" is?

Thanks,
Evan

BTW, here is my working source (ignore the "sig" function; that is my signature function.):

include <iostream>

include <string>

using namespace std;

int main () {
double a, b, c, x, x2, sq;
a = b = c = x = 0;

cout << "Enter a value for coefficient \"a\": > ";
cin >> a;
cout << "Enter a value for coefficient \"b\": > ";
cin >> b;
cout << "Enter a value for coefficient \"c\": > ";
cin >> c;

sq = sqrt (pow(b,2) - (4*a*c));

x = (-b + sq) / 2 * a;
x2 = (-b - sq) / 2* a;

cout << "Here is the first answer: " << x << endl;
cout << "Here is the other answer: " << x2 << endl;

cin.get ();
cin.get ();
return 0;

}

[Ian Walker]

1.#IND is an "indefinite", which means what it says - you are trying to take the real root of a negative number, which is undefined.

When you use the sqrt function, the compiler calls the MSVCRT runtime library, which is documented at http://msdn.microsoft.com/library/en-us/vclib/html/_crt_sqrt.asp

The same runtime library also helpfully provides the _isnan function which checks to see if a number is of this type. So you could modify your result display code thus:

if(_isnan(sq))
{
cout << "No real roots" << endl;
} else {
cout << "Here is the first answer: " << x << endl;
cout << "Here is the other answer: " << x2 << endl;
}

Hope this helps.

Ian

[Stepho]

I think this #IND thing happens when the chosen coefficients lead to calculating the square root of a negative number. Which is not valid when working with "reals".

May be you have mistaken something in the line :
sq = sqrt (pow(b,2) - (4ac));
Or you need to work with complex numbers (I've never done it in C, so I can't help there)

In order to make your code compile, I had to add #include <cmath>. And also, b*b or pow(b,2) gives the same results.