dealing with some logic errors

[raisudessu]

Hello peeps! I'm having some problems with my codes. I had done a quadratic equation calculator programm. The problems were some of the number that I key in will result in appearing "nan" and also the other one would be no output printed at all. I hope that anybody can explain to me about how to deal with this kind of logic error.

[code]
#include <iostream>
#include <cmath>
using namespace std;

int main()
{
double x1,x2,x3,x4,x5,x6,sum1;
int a,b,c;
cout << " ax^2+bx+c " << endl;
cout << " please enter the value of a,b,c : ";
cin >> a >> b >> c;

if(a==0){cout << " undefined. " << endl;}
else{sum1 = (b*b)-(4*a*c);}
x1 = ((-b+sqrt(sum1))/(2*a))*1.0;
x2 = ((-b-sqrt(sum1))/(2*a))*1.0;
if(sum1<0){
cout << " there will be two complex roots. " << endl;
cout << " the roots are " << x1 << " , " << x2 << endl;
}
else if(sum1=0){
cout << " there will be equal roots. " << endl;
cout << " the roots are " << x1 << " , " << x2 << endl;
}
else if(sum1>0){
cout << " the will be two real roots. " << endl;
cout << " the roots are " << x1 << " , " << x2 << endl;
}
return 0;
}

[lastchance]

Please use code tags - it will make it much easier for people to read your code and help you.

There are a number of issues with your code - and these are noted without actually compiling and running.

If a == 0 you correctly point out that the result is undefined (because the equation won't actually be quadratic); however, you then keep going anyway, and so you will fail because of dividing by 0 when you try to calculate x1 and x2.

If sum1 < 0 then you will find that taking its square root is not a clever idea (actually, it will automatically crash) - one way or another you will have to work with complex numbers at this point.

Your condition (sum1 = 0) doesn't do what you think it does: single = means assignment (i.e. giving a value); logical comparison requires ==.


There are other issues. Making a, b, c integers is possible, but rather restrictive.
The *1.0 in the formulae for x1 and x2 is pointless: sum1 is double, so the result of the previous bracketed calculation will be double. sum1 isn't a sensible name: it clearly isn't a sum: most people call it a discriminant (because it discriminates between the possible types of outcome).


I suggest that you modify your code in line with some or all of these suggestions and repost with code tags. Give examples of the cases which fail.