Solve Floating Point Error: 4th Order Runge-Kutta Method

[Vagrant]

This is a program to slove differential equation using fourth order Runge-Kutta method. The question is dy/dx=x²+y²
when y(0)=0, h=0.2
find y(0.4) (Ans= 0.021360)

#include<stdio.h>
#include<conio.h>
float f(float x, float y)
{
return (x*x+y*y);
}
void main()
{
float x1,y1,x2,y2,h=0.2,m1,m2,m3,m4;
clrscr();
printf("Enter the initial values of x and y\n");
scanf("%f%f",&x1,&y1);
while(1)
{
m1=f(x1,y1);
m2=f(x1+h/2,y1+(m1*h)/2);
m3=f(x1+h/2,y1+(m2*h)/2);
m4=f(x1+h,y1+m3*h);
y2=y1+(h/6)*(m1+(2*m2)+(2*m3)+m4);
x2=x1+h;
if(x2==0.4)
break;
else
{
x1=x2;
y1=y2;
}
}
printf("The value of y at x= 0.4 is %f\n",y2);
getch();
}

However this program is not working and giving the message Abnormal value: Floating point overflow. Can anyone please find out the error in this?
Thanks

 

[gendou2]

right away i notice that you will never get out of that while(1) loop unless x2==0.4
is that guaranteed to happen for every possible input?
if not, you've got an infinite loop. bad.

 

[Vagrant]

No, it will only be satisfied for this particular question, i.e. initial value of x=0, h=0.2.
It's not a very good program but it should work at least for this case, which it is not doing.

 

[Borek]

These are float values, chances that x2 will be EXACTLY 0.4 are slim.

You should get information about the line where the error occurs. If you don't have a debugger that will let you trace code step by step, put something like printf("x2 = %f\n") into code to see how x2 values change. You may check every variable this way.

 

[Vagrant]

I did that. The loop is not stopping after x2=0.4, why is break not working?

 

[Borek]

Try if (x>=0.4).

 

[Vagrant]

Yes, it works now.

These are float values, chances that x2 will be EXACTLY 0.4 are slim.

Was this the problem?

Thank you.

 

[Borek]

Was this the problem?

As you see. Generally you can't compare floats with x==y, rather with abs(x-y)<treshold.

 

[Vagrant]

I'll keep this in mind from now on. Thank you for the time and effort you put into this.