@vishal: Floats and doubles are stored in different formats, so
looking at the first 8 hex digits of the two numbers isn't really
helpful.
The expression f < 0.8 is evaluated as (double)f < 0.8, so it would be
more useful to print all 16 hex digits of (double)0.8f and 0.8. Then
it would be easy to see that 0.08f < 0.08.
Dave
On May 29, 11:35 pm, Vishal Thanki <vishaltha...@gmail.com> wrote:
> you may want to check how the floats and doubles are stored into
> memory using ieee notation.
>
> i tried to print 0.08 and 0.08f in hex format and got the following result.
>
> vishal@ubuntu:~/progs/c\ 10:03:56 AM >$ cat fl.c
> #include <stdio.h>
> int main()
> {
> float f=0.08;
> if (f < 0.08f)
> printf("hi\n");
> else
> printf("hello\n");
>
> printf ("%x %x\n",0.08, 0.08f);
> return 0;}
>
> vishal@ubuntu:~/progs/c\ 10:04:06 AM >$ gcc fl.c
> fl.c: In function ‘main’:
> fl.c:10: warning: format ‘%x’ expects type ‘unsigned int’, but
> argument 2 has type ‘double’
> fl.c:10: warning: format ‘%x’ expects type ‘unsigned int’, but
> argument 3 has type ‘double’
> vishal@ubuntu:~/progs/c\ 10:04:11 AM >$ ./a.out
> hello
> 47ae147b 3fb47ae1
> vishal@ubuntu:~/progs/c\ 10:04:14 AM >$
>
> ps: please ignore the warning in the code.
>
> On Sun, May 29, 2011 at 10:23 PM, sravanreddy001
>
>
>
> <sravanreddy...@gmail.com> wrote:
> > and, I read it long time back that.. the value of 0.8 alone will be stored
> > as 0.799999999995 (not sure on the number of 9's but.. the last digit in the
> > precision will be a 5)
> > that could be a reason.
> > may be what "vishal" said is correct.
>
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