Display integer

[Binarydude87]

With the loop below, is there a way to display the actual number without the leading zeros (scientific notation) or will it just display 0 since there are so many leading zeros?

num = 1;

while (num > 0)
{
    num /= 2;
}

cout << num;

[Peter87]

You loop until num becomes zero so that is the value that num will have after the loop.

[Duthomhas]

Perhaps we could help better if we understood what you are trying to do with this code.

[Binarydude87]

Well the book I am studying from asks to write a short program to determine the smallest number, x_min, used on the computer you are using. Note that your computer will be unable to reliably distinguish between zero and a quantity that is smaller than this number.

Maybe I am going about it the wrong way

[TwilightSpectre]

#include <cfloat>
#include <iostream>

int main() {
    std::cout << "Minimum float value: " << FLT_MIN << "\n";
    std::cout << "Minimum double value: " << DBL_MIN << "\n";
    std::cout << "Minimum long double value: " << LDBL_MIN << "\n";

    return 0;
}

Is that what you wanted? If you wanted the integral types, you can use climits as well.

[JLBorges]

> determine the smallest number, x_min, used on the computer you are using.
> Note that your computer will be unable to reliably distinguish between zero
> and a quantity that is smaller than this number.

Determining an approximate answer computationally is not too difficult.

#include <iostream>
#include <cfloat>
#include <iomanip>

int main()
{
    std::cout << std::scientific << std::setprecision(20) 
              << "actual epsilon: " << DBL_EPSILON << '\n' ;

    double min_last = 10 ;

    for( double d = 1 ; d > 0.0001 ; d -= 0.0000001 )
    {
        double v = d ;

        double last_seen ;
        while( v > DBL_EPSILON )
        {
            last_seen = v ;
            v /= 2 ;
        }

        if( last_seen < min_last ) min_last = last_seen ;
    }

    std::cout << "  minimum seen: " << min_last << '\n' ;
}

http://coliru.stacked-crooked.com/a/5c08f980ac093d06

[long double main]

Note that your computer will be unable to reliably distinguish between zero and a quantity that is smaller than this number.

So, if I understand you correctly, are you looking for the smallest representable number greater than zero?

If so, you can read this:
http://stackoverflow.com/a/13699489

Also, this code gives me a pretty small value:

1 2 3 4 5 6 7 8 9 10 11

#include <iostream> int main() { double val = 1.0; while (val > 0) { val /= 2; std::cout << val << '\n'; } }

0.5 0.25 0.125 ... 1.97626e-323 9.88131e-324 4.94066e-324 0

Look at the second-to-last line of the output.
(I can't seem to figure out a way to reliably give me just that value)