Programming Forums > Application Development > C

sixstringartist · Feb 3rd, 2008, 4:41 PM

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.

(Toggle Plain Text)

#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:

(Toggle Plain Text)

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.

sixstringartist · Feb 3rd, 2008, 5:29 PM

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.

Sane · Feb 3rd, 2008, 6:23 PM

(Toggle Plain Text)

    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.

 C Syntax (Toggle Plain Text) 
#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;
 
}
#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;
    
}

sixstringartist · Feb 3rd, 2008, 11:13 PM

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.

(Toggle Plain Text)

#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:

(Toggle Plain Text)

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.

grumpy · Feb 4th, 2008, 2:29 AM

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.

Feb 3rd, 2008, 4:41 PM	#1
sixstringartist Programmer Join Date: Jun 2005 Posts: 68 Rep Power: 4	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. (Toggle Plain Text) #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 + fh; printf("y(%f) = %f \n", t,y); f = 3 + t - y; printf("f(%f,%f) = %f\n", t, y, f); } } return 0; } #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 + fh; 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: (Toggle Plain Text) 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 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.

Feb 3rd, 2008, 5:29 PM	#2
sixstringartist Programmer Join Date: Jun 2005 Posts: 68 Rep Power: 4	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.

Feb 3rd, 2008, 6:23 PM	#3
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 1,869 Rep Power: 5	Re: Simple program odd results (Toggle Plain Text) for(n = 0; n <= 1/h; n++) { printf("n = %f\n", n); n = n+1; 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. C Syntax (Toggle Plain Text) #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; } #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; }

Feb 3rd, 2008, 11:13 PM	#4
sixstringartist Programmer Join Date: Jun 2005 Posts: 68 Rep Power: 4	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. (Toggle Plain Text) #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; } #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: (Toggle Plain Text) 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 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.

Feb 4th, 2008, 2:29 AM	#5
grumpy Programming Guru Join Date: Jun 2005 Location: Adelaide, South Australia Posts: 1,222 Rep Power: 5	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.

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Programming Forums > Application Development > C

Simple program odd results