@Bittu: I wrote that a google years ~= 10^116.5 nanoseconds. If you
take the base-10 logarithm of your time t, you will get about 116.5.
Regarding your code, the dynamic range of doubles exceeds 10^116.5, so
you can calculate n. However, since n is approximately 2^387, but only
the high-order 52 bits is stored, it follows that 1 is insignificant
to n; therefore, subtracting 1 (n--) does not change the numerical
value of n. Your code will run forever.
Dave
On May 13, 4:57 am, bittu <shashank7andr...@gmail.com> wrote:
> @Dave... I think 1 Googol Year is =10^100 not 10^116.5 ?? why u have
> used
>
> so then we have to write the single line program that googol years of
> time ?? & we have processor that can execute the instruction in 10^9
> per second so the time required by googol year in second
> which is equals to time t=pow(10,109)*365*86400 sec.
>
> so program is like
>
> #include <stdio.h>
> #include <math.h>
>
> int main() {
>
> double n = pow(10, 109) * 365 * 86400;
>
> while(n--);
>
> }
>
> Correct me if anything wrong???
>
> Thanks & Regards
> Shashank Mani>>"The Best Way To Escape From The problem is Solve It"
> Computer Science & Engg.
> BIT Mesra
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