floating point division in C language

Question

#include <stdio.h>
#include <conio.h>
#include <math.h>   
float t, delt = 0.004000,n;

printf("enter the value for t=");
scanf("%f", &t);
n = t/ delt;
printf("n is %f",n);
getch();

output: enter the value for t=1 n is 249.999985

I am expecting 250. What am I doing wrong??

Perhaps SO should put a link to [this](http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html) page as a sticky at the top — IllusiveBrian, Sep 24 '14 at 01:57
SO should automatically look for dups before posting. Like bugzilla does — thang, Sep 24 '14 at 01:58

score 3 · Answer 1 · answered Sep 24 '14 at 01:57

You are up against the all-too-common problem that floating point representations are approximations. You may wish to do one of the following:

Round your answer using an unbiased rounding strategy.
Truncate your answer to zero decimal places before printing it (using something like %.0f).
Use an arbitrary-precision arithmetic library such as GMP.

Which one of these is sensible is entirely dependent on the constraints of your application.

floating point division in C language

1 Answers1