Hi Shubham,
This may be because your final data type you have chosen for your output
is float, n the one which you are trying to print is of int data type (i.e. %d).
Suggestion : Try changing %d to %f and see whether it works for you or not?
Thanks Regards,
Pratts
On 01-Mar-2013, at 1:11,
The output of the below code is also zero ,
key reason is floating point number are stored as *mantissa, exponent. *
float dec=1.00;
printf(\n dec=%d , float =%f,dec,dec);
On Fri, Mar 1, 2013 at 12:05 PM, Karthikeyan V.B kartmu...@gmail.comwrote:
O/p will not be 0.
1.00 is the result
on my system every time o/p is 0
using ubuntu 10.04 ,gcc compiler
On Mon, Mar 4, 2013 at 7:34 AM, rohit jangid rohit.nsi...@gmail.com wrote:
output for me for the previous snippet
localhost:slingshot rohitjangid$ ./a.out
1799476872
1799474584
localhost:slingshot rohitjangid$ ./a.out
yup , it is showing
0
0
on ideone as well . so my gcc compiler is i686-apple-darwin11-llvm-gcc-4.2.
that can be the reason . from here it appears 0 is just a coincidence and
it depends on compiler implementation . C doesn't define any such behavior.
On Mon, Mar 4, 2013 at 3:45 PM, Shubham
Ya, its looking like the problem of 'i686-apple-darwin11-llvm-gcc-4.2'.
For me as well it shows different outputs.
On Mon, Mar 4, 2013 at 4:45 PM, rohit jangid rohit.nsi...@gmail.com wrote:
yup , it is showing
0
0
on ideone as well . so my gcc compiler
is i686-apple-darwin11-llvm-gcc-4.2.
yeah true . one interesting thing I noticed is that if you run this code
#includestdio.h
int main()
{
int i = 0;
do {
printf (%d\n,(float)1);
}while(i++ 1);
return 0;
}
one would expect same output in both the rows but surprisingly it came
different for me every time .
output for me for the previous snippet
localhost:slingshot rohitjangid$ ./a.out
1799476872
1799474584
localhost:slingshot rohitjangid$ ./a.out
1710327432
1710325144
localhost:slingshot rohitjangid$ ./a.out
1856128648
1856126360
localhost:slingshot rohitjangid$ ./a.out
1724065416
1724063128
On
code snippet:
*int main()
{
printf (%d\n,(float)((int)(3.5/2)));
return 0;
}*
--
Regards,
SHUBHAM SANDEEP
IT 3rd yr.
NIT ALD.
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I think this is because of type mismatch. You are enforcing your program to
read a floating point number in the way of reading a integer. And they have
totally different format. If you have -Wall turned on, you should see a
warning.
Yanan Cao
On Thu, Feb 28, 2013 at 1:41 PM, Shubham Sandeep
thank you for pointing out that format was the key point.
On Fri, Mar 1, 2013 at 1:19 AM, gmagog...@gmail.com gmagog...@gmail.comwrote:
I think this is because of type mismatch. You are enforcing your program
to read a floating point number in the way of reading a integer. And they
have
O/p will not be 0.
1.00 is the result which when read as %d takes the decimal value of
float 1.00 stored in memory - it will not be 1.00 or 0.
Since float is not stored as direct binary in memory as integer is stored,
instead there's a separate procedure for storing float as binary
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