Re: converting floating point types round off error ....

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Mon, 6 Oct 2008 02:55:12 -0700 (PDT)
Message-ID:
On Oct 5, 1:13 am, ma740988 <ma740...@gmail.com> wrote:

Consider the equation (flight dynamics stuff):

Yaw (Degrees) = Azimuth Angle(Radians) * 180 (Degrees) /

There's a valid reason to use single precision floating point
types.
The number of decimal digits guaranteed to be correct on my
implementation is 6. (i.e numeric_limits < float >::digits10 = 6 )

I'm not quite sure what you mean by "number of decimal digits
guaranteed to be correct". Correct compared to what. To
represent an IEEE floating point exactly in decimal, you need
something like 24 digits. Typically, however, there's
absolutely no need to represent it exactly.

What numeric_limits<>::digits10 guarantees is that any decimal
number with that many digits, converted to the floating point
type and back to decimal with the same number of digits, will
result in the same decimal number. (Indirectly, this also
guarantees that two different decimal numbers with no more
digits will result in two different floating point numbers in
the machine.)

If I'm reading the IEEE standard, I'd could paraphrase the
issue surrounding conversion to a string and back _without_
loss of precision as follows:

If a float is correct to a decimal string with a least 6
significant decimal digits, and then converted back to a
float, then the final number must match the original.

IOW: given
float a = 1. F ;
float aa = 0. ;
std::stringstream s ;
s. precision ( 6 ) ;
s << std::scientific << a ;
s >> aa;
assert ( a != aa ) ;

Except that numeric_limits<>::digits10 doesn't make any
guarantees about converting to a string and back. For this, you
need numeric_limits<>::max_digits10, which will only be
available in the next version of the standard. (For an IEEE
float, the value is 9.)

Note that even a little bit of reasoning will reveal that 6
isn't enough. The mantissa of an IEEE floating point is 24
bits; since the high order bit is always 1, this means that it
can take on 2^23 different values. since 2^23 > 10^6, quite
clearly some different values will map to the same 6 digit
decimal value.

No sweat

I have to serialize the Yaw answer above. The question: Is it
safe to state that my PI representation is useless beyond six
significant digits?

Certainly not. Nine digits may be sufficient, however.

--
James Kanze (GABI Software) email:james.kanze@gmail.com
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