float value

hi all

Code:

#include <stdio.h>
main()
{
float i,j=100.0,k=19456.0;
i=j/k;
printf(******** %f \n",i);
}

output is
******** 0.005139803

i would like round figure it like
0.005.

how can i do this

greetings munna

You can control the precision by trying something like : "%.af" where 'a' is the number of digits following the decimal point.

also you should learn to use int main()
and properly indent your code

Looking at your code, I am guessing you must be using a compiler like TC or BC. Please switch over th GCC. It will help you in the long run.

Originally Posted by kris.c

You can control the precision by trying something like : "%.af" where 'a' is the number of digits following the decimal point.

sorry. it is not working..

Ofcourse it does,

Code:

 # include<stdio.h>
int main()
{
        float i,j=100.0,k=19456.0;
        i=j/k;
        printf("******** &#37;.3f \n",i);
        printf("******** %.5f \n",i);
return 0;
}

my output:

Code:

******** 0.005
******** 0.00514

Originally Posted by kris.c

Looking at your code, I am guessing you must be using a compiler like TC or BC. Please switch over th GCC. It will help you in the long run.

What the heck leads you to guess that?