Simple C program

J_K9 · 10-24-2005, 06:03 AM

Hi,

I'm currently going through this C tutorial, and here's the task I've been set as an exercise:

Quote:

2. Write a program that counts from 1 to 12 and prints the count and its inversion to 5 decimal
places for each count. This will require a floating point number.
1 1.00000
2 .50000
3 .33333
4 .25000
etc.

I thought it'd be fairly easy at first, but for some unknown reason my code doesn't work! Here it is:

Code:

#include <stdio.h>

int main()
{
  int count;
  float invert;
  for(count = 1; count < 13; count++) {
    invert = (1 / count);
    printf("%d\t\t%8.5f\n", count, invert); }
  return 0;
}

The output in terminal is this:

Code:

1		 1.00000
2		 0.00000
3		 0.00000
4		 0.00000
5		 0.00000
6		 0.00000
7		 0.00000
8		 0.00000
9		 0.00000
10		 0.00000
11		 0.00000
12		 0.00000

I'm not sure why this is happening, because it is a float, and I've set the output to have 5 decimal places. Any help would be appreciated.

M.

dmail · 10-24-2005, 06:22 AM

its quite simple when you know whats going on.
this line
invert = (1 / count);
if you replace the variables and constant with there types
float = int/int; do you see the error now?

J_K9 · 10-24-2005, 11:36 AM

Of course! Thank you so much!

For those who might have had similar problems, just replace the 'invert = (1 / count);' with 'invert = (1.00 / count);'