From:

Pete Becker <pete@versatilecoding.com>

Newsgroups:

comp.lang.c++

Date:

Tue, 06 Oct 2009 11:06:45 -0400

Message-ID:

<f8mdndBqTPSYwFbXnZ2dnUVZ_sSdnZ2d@giganews.com>

On 6 Okt, 15:04, Victor Bazarov <v.Abaza...@comAcast.net> wrote:

No. There are infinitely many real numbers between

any two consecutive FP numbers.

Rune Allnor wrote:

Or 0.25, 0.125, 0.0625, etc., or *any combination thereof*. Then factor

in the exponents.

That's one of the few

*Few*? You're kidding, of course, aren't you?

On 6 Okt, 14:22, Nick Keighley <nick_keighley_nos...@hotmail.com>

wrote:

wrote:

On 6 Oct, 12:39, Rune Allnor <all...@tele.ntnu.no> wrote:

try 0.5

On 6 Okt, 11:16, Carsten Fuchs <CarstenFu...@T-Online.de> wrote:

You should choose between *either* human readability *or*

data integrity. You can't have both.

To demonstrate the point, try

-----------

#include <iostream>

int main()

{

float a=0.3;

double b = a;

std::cout << "a = " << a << std::endl;

std::cout.precision(12);

std::cout << "b = " << b << std::endl;

return 0;}

-----------

Output:

a = 0.3

b = 0.300000011921

There are two problems here:

1) Numbers that are exact in decimal notation

have no exact floating-point representation,

only an approximation.

I would like to serialize a float f1 to a string s, then unserialize s back to a float f2

again, such that:

* s is minimal (use only the precision that is required)

and preferably in decimal notation,

* f1==f2 (exact same value after the roundtrip)

(The first property is for human readers and file size, the second is for data integrity.)

again, such that:

* s is minimal (use only the precision that is required)

and preferably in decimal notation,

* f1==f2 (exact same value after the roundtrip)

(The first property is for human readers and file size, the second is for data integrity.)

You should choose between *either* human readability *or*

data integrity. You can't have both.

To demonstrate the point, try

-----------

#include <iostream>

int main()

{

float a=0.3;

double b = a;

std::cout << "a = " << a << std::endl;

std::cout.precision(12);

std::cout << "b = " << b << std::endl;

return 0;}

-----------

Output:

a = 0.3

b = 0.300000011921

There are two problems here:

1) Numbers that are exact in decimal notation

have no exact floating-point representation,

only an approximation.

try 0.5

Or 0.25, 0.125, 0.0625, etc., or *any combination thereof*. Then factor

in the exponents.

:-)

That's one of the few

*Few*? You're kidding, of course, aren't you?

No. There are infinitely many real numbers between

any two consecutive FP numbers.

That's a different issue. The question was how to write out values that

are represented in floating-point so that you can read them back in exactly.

Using the IEEE

definition of "single precision float", there are *about* (2^32 - 2^24)

decimal values that can be represented *exactly*. Each of all the other

values (infinite number of them, of course) are rounded to one of the

*more than three billion* representations (my arithmetic may be off a tad).

definition of "single precision float", there are *about* (2^32 - 2^24)

decimal values that can be represented *exactly*. Each of all the other

values (infinite number of them, of course) are rounded to one of the

*more than three billion* representations (my arithmetic may be off a tad).

Three billions is still a finite number of representations.

True, it is large enough to be useful, but still finite.

But there are only a finite number of floating-point values, so being

finite is not, in itself, a limitation.

> decimal numbers that can be represented

No, he is not "in for trouble". He just needs to realise that computer

representation of the floating point numbers have limitations, and *stay

within those limitations*.

exactly on binary format. In fact, decimal numbers on the form

x = 2^n

where n is selected from a certain subset of integers, can be

represented exactly, as binary numbers. If the OP can accept

such a constraint on the decimal numbers he wants to work with,

then by all means, disregard what I said. But if he wants to

work with arbitrary numbers, he is in for trouble.

x = 2^n

where n is selected from a certain subset of integers, can be

represented exactly, as binary numbers. If the OP can accept

such a constraint on the decimal numbers he wants to work with,

then by all means, disregard what I said. But if he wants to

work with arbitrary numbers, he is in for trouble.

No, he is not "in for trouble". He just needs to realise that computer

representation of the floating point numbers have limitations, and *stay

within those limitations*.

As I understand the question, the OP wants to break out

of those limitations: Exact conversions between base-10

and base-2 numbers, eliminating approximation errors etc.

No, he's looking for conversions of the finite set of values that can be

represented as floating-point.

--

Pete

Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of

"The Standard C++ Library Extensions: a Tutorial and Reference"

(www.petebecker.com/tr1book)

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