# Re: rounding errors

```"Dominic R. Scales" <Dominic.Scales@aerosensing.de> writes:

>HELP!

>  What gives? Is there any numerical math guy/gal out there
>  who can tell me how this happens? It seems to me, that
>  the accuracy of the second/third cast ist WAY off.

>   a=double('2.56989')
>   b=double( 2.56989 )
>   c=double(float('2.56989'))

>   print,a,b,c,format='(d)'

>       2.5698900000000000 <---- this is what i want to have
>       2.5698900222778320
>       2.5698900222778320

The first question that comes into my mind is why don't you simply cast your
literal as double precision in the first place:

IDL> a = 2.56989d0
IDL> print,a,format='(d)'
2.5698900000000000

It's instructive to look at these numbers in the binary representation actually
used by the computer.  The floating point number 2.56989 has the following
representations in hexadecimal and binary formats:

Hex:     40247914
Binary:  01000000001001000111100100010100
^^^^^^^^^^^^^^^^^^^^^^

when this is converted from floating point to double precision, as in your
examples b and c above, you get

Hex:     40048F2280000000
Binary:  0100000000000100100011110010001010000000000000000000000000000000
^^^^^^^^^^^^^^^^^^^^^^

It's a little harder to tell in hex format, but the binary format makes it
plain that the mantissa part is exactly the same as in the original floating
point number, just shifted over a little (the exponent part is bigger in double
precision), and filled with zeroes, just as you would expect.  The confusion
results from the distinction between decimal representation used by people and
the binary representation used by computers.

William Thompson
```