Programming Forums - Simple program odd results

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Simple program odd results

Ive been out of coding practice for several months now so my debugging is rusty. I cant seem to figure out why Im getting the results that I am. Its a simple program used to aid my calculations in a diff. eq. course.

#include<stdio.h>

#include<math.h>





int main()

{

  float y;

  float f;

  float h;

  int n = 0;

  float t;

  while(1){

    printf("Enter starting time: \n");

    scanf("%f", &t);

    printf("Enter starting y value: \n");

    scanf("%f", &y);

    printf("Enter time step(h): \n");

    scanf("%f", &h);



    f = 3 + t - y;

    printf("y(%f) = %f\n", t, y);

    printf("f(%f,%f) = %f\n", t, y, f);



    for(n = 0; n <= 1/h; n++)

    {

       printf("n = %f\n", n);

       n = n+1;

       printf("n = %f\n", n);

       t = t + h;

       y = y + f*h;

       printf("y(%f) = %f \n", t,y); 

       f = 3 + t - y;

       printf("f(%f,%f) = %f\n", t, y, f);

     }

  }

  return 0;

}

and the results I get for t = 0, y = 1, h = 0.05 are:

Enter starting time:

0

Enter starting y value:

1

Enter time step(h):

0.05

y(0.000000) = 1.000000

f(0.000000,1.000000) = 2.000000

n = 0.000000

n = 0.000000

y(0.050000) = 1.100000

f(0.050000,1.100000) = 1.950000

n = 0.050000

n = 0.050000

y(0.100000) = 1.197500

f(0.100000,1.197500) = 1.902500

n = 0.100000

n = 0.100000

y(0.150000) = 1.292625

f(0.150000,1.292625) = 1.857375

n = 0.150000

n = 0.150000

y(0.200000) = 1.385494

f(0.200000,1.385494) = 1.814506

n = 0.200000

n = 0.200000

y(0.250000) = 1.476219

f(0.250000,1.476219) = 1.773781

n = 0.250000

n = 0.250000

y(0.300000) = 1.564908

f(0.300000,1.564908) = 1.735092

n = 0.300000

n = 0.300000

y(0.350000) = 1.651663

f(0.350000,1.651663) = 1.698337

n = 0.350000

n = 0.350000

y(0.400000) = 1.736580

f(0.400000,1.736580) = 1.663420

n = 0.400000

n = 0.400000

y(0.450000) = 1.819751

f(0.450000,1.819751) = 1.630250

n = 0.450000

n = 0.450000

y(0.500000) = 1.901263

f(0.500000,1.901263) = 1.598737

The for loop seems to be the root of the problem, and I wasnt getting expected behavior from the "n" variable so I tried to brute force a behavior (n = n + 1) but I still get back the previous value for n, and n seems to always increase by h no matter what I specify in the loop. I know this is a very trivial problem, but I cant figure out what the bug is.

Thanks for the help.

Re: Simple program odd results

My window for editing has expired and I thought I should add that Im getting correct approximations independant of the step size used. "n" just seems to not be playing nicely.

Re: Simple program odd results

    for(n = 0; n <= 1/h; n++)

    {

       printf("n = %f\n", n);

       n = n+1;

Why are you incrementing n twice in the same for loop? The n++ in the loop declaration performs the n = n+1 for you.

Edit: And as a side note. The <= comparator for an integer and float might not behave exactly as you expect. Look at the following sample. It says that 20 is not <= 20.0. But change n to 19, and you get True.

#include <stdio.h>
 
int main() {
 
    float f = 0.05;
    int n = 20;
 
    printf("%f\n", 1/f);
    printf("%d\n", n);
    printf("%s\n", n <= 1/f ? "True" : "False");
    printf("\n");
 
    return 0;
 
}

Re: Simple program odd results

yes, using a float in the condition of a for loop isnt a good idea, but Im not sure if thats really the problem since it does execute the correct number of loops. I did discover that the printf statement of n has something to do with the odd results. Since its declared as an int and I try to print a float bad things happened and I would appreciate some insight into why.

#include<stdio.h>





int main()

{

  int n = 0;

  int i = 0;



  for(i = 0; i < 10; i++)

  {

    printf("n = %f\n", n);

    n += 1;

  }

  return 0;

}

Outputs:

n = -0.025121

n = -0.025121

n = -0.025121

n = -0.025121

n = -0.025121

n = -0.025121

n = -0.025121

n = -0.025121

n = -0.025121

n = -0.025121

but if I change the %f in the print statement to a %d I get the expected results.

Re: Simple program odd results

Your results are hardly surprising. You are attempting to print n out, but the %f format specifier tells printf() that n is a (double) floating point value. printf() will not magically convert n to a floating point value; it will print out the floating point value that happens to be represented with a double containing the same set of bits as the int named n.

Formally, passing an argument to printf() that doesn't match the corresponding format (eg using %f for an int) yields undefined behaviour.