"Vladimir S. Oka" <novine@btopenworld.com> writes:[color=blue]
> ramu wrote:[color=green]
>> main()
>> {
>> union {
>> float i;
>> int j;
>> } un;
>>
>> un.i=2.35;
>> printf("%f\n",un.i);
>>
>> un.j=3;
>> printf("%d\n",un.j);
>>
>> printf("%f\n",un.i);
>> }
>>
>> Am getting output as:
>> 2.350000
>> 3
>> 0.000000
>>
>> I wonder how it gives 0.000000 for the value of un.i when i printed
>> its value for the second time. can anyone help me out?[/color]
>
> In short (there was a more detailed discussion recently), all members of
> an union share storage. If you think in terms of memory, they are all
> stored in the same memory area (large enough to hold the largest union
> member). Writing a value into one member, and then accessing another
> will give implementation dependant results. On your implementation, it
> seems that when you read a piece of storage, with integer 3 stored in
> it, as a float, it happens you get a float 0.00000. You may have ended
> up with an illegal value of the float as well, and run into all sorts
> of problems. You were just lucky 9or knew _exactly_ waht you were
> doing).[/color]
Actually, the value is (apparently) 0.00000 within the precision
displayed by printf's "%f" format. It's probably not exactly zero.
You might try using "%g" -- or "%.20g" if you want to see even more
digits.
Also, you need to add "#include <stdio.h>" (required if you're using
printf()), "main()" should be "int main(void)", and you should have a
"return 0;" at the end.
--
Keith Thompson (The_Other_Keith)
kst-u@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.
