dev-cpp-users

[Dev-C++] Question about bit representation

[Ioannis V. <iv...@em...>]

In a newsgroup out there there was a girl who wanted to see the
representation of a given double in her system. All others provided low
level messy stuff involving & operators etc.


I provided the code:


#include <iostream>
#include <bitset>
#include <limits>


int main()
{
 using namespace std;

 double d=3D1;

 unsigned char *p=3Dreinterpret_cast<unsigned char *>(&d);

// There are systems where 1 byte is more than 8 bits
 bitset<numeric_limits<unsigned char>::digits> byte;

 for(unsigned i=3D0; i<sizeof(d); ++i)
 {
   byte=3Dp[i];

   for(unsigned j=3D0; j<byte.size(); ++j)
      cout<<byte[j];=20
 }
=20
 cout<<endl;
}


However it seems to be printed in the opposite way (try making d an =
unsigned
char). Perhaps it is the way it is implemented in the system, or I am =
taking
something in the opposite way that I can't find out?






Ioannis Vranos
=20
* Programming pages: http://www.noicys.freeurl.com
* Alternative URL 1: http://run.to/noicys
* Alternative URL 2: http://www.noicys.cjb.net

Re: [Dev-C++] Question about bit representation

[<or...@vp...>]

Ioannis Vranos wrote:
> However it seems to be printed in the opposite way (try making d an unsigned
> char). Perhaps it is the way it is implemented in the system, or I am taking
> something in the opposite way that I can't find out?

It is probably because of endianness, and it is normal. Because you 
directly access the memory, where the double is stored, it is 
system-dependent whether the low or high order byte comes first.

Most systems are little endian (but there are many that are big endian). 
Look on the net for more details about little and big endian stuff.
--
Greetings,
   Balázs

Re: [Dev-C++] Question about bit representation

[Per W. <pw...@ia...>]

I'm not sure if a byte by definition may be anything but 8 bit.

However, at least IBM have old irons with 9-bit characters. I assume that
they are using own-defined floating point data too.

For just about all machines in the world, the fp format is IEEE.

A double is then stored as

S EEEEEEEEEEE FFFFFFFFFFF.....FFFFF
0 1..11       12..63

E=2047, F!=0 -> Not a number
E=2047, F=0, S=1 -> -infinity
E=2047, F=0, S=0 -> +infinity
E!=0 -> normalized number -1^S * 2^(E-1023) * (1.F)
E=0, F!=0 -> unnormalized number -1^S * 2^(-1022) * 0.F
E=0, F=0, S=1 -> -0
E=0, F=0, S=0 -> +0

Similar for floating point, but
8-bit E, 23-bit F

Replace E=2047 with E=255
Replace (E-1023) with (E-127)
Replace 2^(-1022) with 2^(-126)

/Per W

On Mon, 7 Jul 2003, Ioannis Vranos wrote:

> In a newsgroup out there there was a girl who wanted to see the
> representation of a given double in her system. All others provided low
> level messy stuff involving & operators etc.
>
>
> I provided the code:
>
>
> #include <iostream>
> #include <bitset>
> #include <limits>
>
>
> int main()
> {
>  using namespace std;
>
>  double d=1;
>
>  unsigned char *p=reinterpret_cast<unsigned char *>(&d);
>
> // There are systems where 1 byte is more than 8 bits
>  bitset<numeric_limits<unsigned char>::digits> byte;
>
>  for(unsigned i=0; i<sizeof(d); ++i)
>  {
>    byte=p[i];
>
>    for(unsigned j=0; j<byte.size(); ++j)
>       cout<<byte[j];
>  }
>
>  cout<<endl;
> }
>
>
> However it seems to be printed in the opposite way (try making d an unsigned
> char). Perhaps it is the way it is implemented in the system, or I am taking
> something in the opposite way that I can't find out?
>
>
>
>
>
>
> Ioannis Vranos
>
> * Programming pages: http://www.noicys.freeurl.com
> * Alternative URL 1: http://run.to/noicys
> * Alternative URL 2: http://www.noicys.cjb.net
>
>
>
>
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