problems with sqrt function

Dear All,

I am terribly confused by something at the moment. If anyone could tell me what they think the problem is that would be great.

Code:

---


#include<stdio.h>
#include<math.h>

int main()
{
    double a=(sqrt(4*M_PI*M_PI+1)/2),b=(sqrt(M_PI*M_PI+1/4));
   
    printf("a=%f\t\tb=%f",a,b);
   
    return(0);
}

---

As you can easily tell mathematically my definitions of a and b are identical. However when one runs the program a has the correct value stored, but b has the value M_PI.

Thank you for all of your time.
Edz

This is because this is computers, not mathematics!

C believes that 1/4 == 0, since it does integer division when dividing integers.

1/4 = 0.

try this way

Code:

---

    double a=(sqrt(4.0*M_PI*M_PI+1.0)/2.0),b=(sqrt(M_PI*M_PI+1.0/4.0));

---

Kurt

Thank you all

you can also type cast by using the data type a constant is referring to..

like:
int(1.0)=1
float(1)=1.0

M_PI is nonstandard, although GCC's <math.h> has it. For portability, use something like

Code:

---

  const double Pi = 4.*atan(1.);

---

instead.

Or

Code:

---

#ifndef M_PI
    #define M_PI 3.14159265358979324
#endif

---

(From memory, might want to check the value . . . .)