Value bits as compile-time constant!

[Frederick Gotham]

Over on comp.lang.c, Hallvard B Furuseth devised a compile-time constant
representing the amount of value representation bits in an unsigned
integer type. In true C fashion, here it is as a macro:


/* Number of bits in inttype_MAX, or in any (1<<k)-1 where 0 <= k < 3.2E+
10 */
#define IMAX_BITS(m) ((m) /((m)%0x3fffffffL+1) /0x3fffffffL %0x3fffffffL
*30 \
+ (m)%0x3fffffffL /((m)%31+1)/31%31*5 + 4-12/((m)%
31+3))


You supply it with an unsigned integer value whose bit-pattern consists
of all 1's, e.g.:

1 (Decimal: 1)
11 (Decimal: 3)
111 (Decimal: 7)
11111111 (Decimal: 255)


To find out the amount of value bits in an unsigned integer type, you
supply it with that type's max value (i.e. a bit pattern of all 1's). The
handiest way to get the max value of an unsigned integer type is to
assign -1 to it:

unsigned const max_value = -1;

unsigned const amount_value_bits = IMAX_BITS( max_value );


Here's some template code I've written which makes use of it:


template<class T>
struct IMaxBits {

template<T m>
struct AllOnes {

static unsigned const val =
m /(m%0x3fffffffL+1) /0x3fffffffL %0x3fffffffL *30
+ m%0x3fffffffL /(m%31+1)/31%31*5 + 4-12/(m%31+3);

};

};


template<class T>
struct AmountValueBits {

static unsigned const val = IMaxBits<T>::template AllOnes<-1>::val;

};


#include <cstdlib>
#include <cstring>
#include <cstddef>
#include <cassert>

template<std::size_t width>
inline const char* CenterHoriz( const char * const p )
{
using std::memset;
using std::strlen;
using std::size_t;

static char spaces[width + 1];


memset( spaces, ' ', width );

size_t const len = strlen( p );

assert( width >= len );


char * const pos = spaces + ( width / 2 - len / 2 );


memcpy( pos, p, len );

return spaces;
}



#include <iostream>

extern char const str_uchar[] = "unsigned char";
extern char const str_ushort[] = "unsigned short";
extern char const str_uint[] = "unsigned int";
extern char const str_ulong[] = "unsigned long";

template<class T, const char* str>
void PrintRow()
{
using std::cout;

unsigned const total_bits = sizeof(T) * AmountValueBits<unsigned
char>::val;
unsigned const val_bits = AmountValueBits<T>::val;
unsigned const pad_bits = total_bits - val_bits;

const char * const spaces_val =
val_bits > 99 ? "" : ( val_bits > 9 ? " " : " " );

const char * const spaces_pad =
pad_bits > 99 ? "" : ( pad_bits > 9 ? " " : " " );

const char * const spaces_total =
total_bits > 99 ? "" : ( total_bits > 9 ? " " : " " );


cout << "||" << CenterHoriz<21>(str) << "|| "
<< spaces_val << val_bits << " || " << spaces_pad <<
pad_bits
<< " || " << spaces_total << total_bits << " ||\n"

"---------------------------------------------------------------------
\n";
}

int main()
{
using std::cout;

cout <<

"
==============================================\n"
" || Value bits || Padding Bits || Total Bits
||\n"

"=====================================================================
\n";

PrintRow<unsigned char, str_uchar>();
PrintRow<unsigned short, str_ushort>();
PrintRow<unsigned, str_uint>();
PrintRow<unsigned long, str_ulong>();

cout << "\n\n\n";

std::system( "PAUSE" );
}



I'd be interested to hear if anyone gets some "interesting" values.

 

[Alf P. Steinbach]

* Frederick Gotham:

Over on comp.lang.c, Hallvard B Furuseth devised a compile-time constant
representing the amount of value representation bits in an unsigned
integer type. In true C fashion, here it is as a macro:


/* Number of bits in inttype_MAX, or in any (1<<k)-1 where 0 <= k < 3.2E+
10 */
#define IMAX_BITS(m) ((m) /((m)%0x3fffffffL+1) /0x3fffffffL %0x3fffffffL
*30 \
+ (m)%0x3fffffffL /((m)%31+1)/31%31*5 + 4-12/((m)%
31+3))

In C++ you'd use std::numeric_limits::digits instead.

 

[Howard]

Over on comp.lang.c, Hallvard B Furuseth devised a compile-time constant
representing the amount of value representation bits in an unsigned
integer type. In true C fashion, here it is as a macro:

What does this mean: "the amount of value representation bits"?

-Howard

 

[Victor Bazarov]

What does this mean: "the amount of value representation bits"?

Substitute 'amount' with 'number', does it make it simpler? If not,
substitue 'value representation' with 'value-representing'. If after
that it's not clear, then the definition is probably "the number of
bits involved in representing the value".

V

 

[Frederick Gotham]

Howard posted:

What does this mean: "the amount of value representation bits"?

I'll work off an example.

Let's say that a byte is 8-Bit, and that sizeof(unsigned short) == 4.

Therefore, an "unsigned short" takes up 32 bits in memory.

But that doesn't mean an unsigned short can use 32 bits to store its value.
It might only use 30 of those bits; the rest are padding bits. The macro
yields a compile time constant telling us how many bits take part in the
"value representation" of the unsigned integer type.

 

[Howard]

Howard posted:




I'll work off an example.

Let's say that a byte is 8-Bit, and that sizeof(unsigned short) == 4.

Therefore, an "unsigned short" takes up 32 bits in memory.

But that doesn't mean an unsigned short can use 32 bits to store its
value.
It might only use 30 of those bits; the rest are padding bits. The macro
yields a compile time constant telling us how many bits take part in the
"value representation" of the unsigned integer type.

Interesting. I don't recall ever hearing of padding bits in a stored
integer before (except perhaps way back in my old 60-bit Cyber mainframe
days...but those memories are fading). Do any modern machines actually have
them? All the machines I've worked with in recent years allow you to use
(for example) all 32 bits of a 32-bit integer. The CPU may contain extra
"flag" bits, but those don't reduce the amount useable in stored memory.

-Howard

 

[Frederick Gotham]

Howard posted:

Interesting. I don't recall ever hearing of padding bits in a stored
integer before (except perhaps way back in my old 60-bit Cyber
mainframe days...but those memories are fading).

I think super-computers do it, something to do with "emphasizing
floating-point arithmetic"...

All the machines I've worked with in
recent years allow you to use (for example) all 32 bits of a 32-bit
integer.

All the bits in memory belong to you, and they're yours to play around
with, e.g.:

unsigned short i;

unsigned char *p =
reinterpret_cast<unsigned char*>( &i );

unsigned char * const p_over =
reinterpret_cast<unsigned char*>( &i + 1 );


do
{
*p++ = 0;
} while( p != p_over );


But if you store a value as an unsigned short, then you don't have 2^32
unqiue values, but rather 2^30... even though it takes up 32 bits in
memory. It's all about range really:

0
through
( 2^(amount value bits) ) - 1

 

[Howard]

Howard posted:




I think super-computers do it, something to do with "emphasizing
floating-point arithmetic"...




All the bits in memory belong to you, and they're yours to play around
with, e.g.:

unsigned short i;

unsigned char *p =
reinterpret_cast<unsigned char*>( &i );

unsigned char * const p_over =
reinterpret_cast<unsigned char*>( &i + 1 );


do
{
*p++ = 0;
} while( p != p_over );


But if you store a value as an unsigned short, then you don't have 2^32
unqiue values, but rather 2^30... even though it takes up 32 bits in
memory. It's all about range really:

0
through
( 2^(amount value bits) ) - 1

(Assuming unsigned short is 30 bits, I guess. On my PC, it's 16 bits, and I
think on my Mac as well.)

Thanks,
-Howard

 

[Alf P. Steinbach]

* Howard:

(Assuming unsigned short is 30 bits, I guess. On my PC, it's 16 bits, and I
think on my Mac as well.)

Go back up the thread to find the context for the example.

 

[Jerry Coffin]

[ ... ]

Interesting. I don't recall ever hearing of padding bits in a stored
integer before (except perhaps way back in my old 60-bit Cyber mainframe
days...

Yup -- 40 significant bits and 20 padding bits.

but those memories are fading).

If you ever want to look at it, _Design of a Computer: The Control
Data 6600_, by James Thornton has been scanned in and is available
for download a number of places around the 'net. If you're ever
struck with a _reallY_ serious bout of nostalgia, check out:
http://members.iinet.net.au/~tom-hunter/
the "Desktop Cyber emulator home." It's almost even close to topical,
being written in portable C.

Do any modern machines actually have
them? All the machines I've worked with in recent years allow you to use
(for example) all 32 bits of a 32-bit integer. The CPU may contain extra
"flag" bits, but those don't reduce the amount useable in stored memory.

I've seen a compiler for a 12-bit DSP that stored char as 8
significant bits with 4 padding bits. That was a few years ago
though. In the last few years, DSPs have started to look quite a bit
more like mainstream CPUs (while CPUs have started to look a bit more
like DSPs).

 

[Alf P. Steinbach]

* Jerry Coffin:

I've seen a compiler for a 12-bit DSP that stored char as 8
significant bits with 4 padding bits.

That would be a non-conforming compiler. For char all bits participate
in the value representation. Per the standard.

 

[Jerry Coffin]

* Jerry Coffin:

That would be a non-conforming compiler. For char all bits participate
in the value representation. Per the standard.

I should have pointed out that this was a C (89/90) compiler, which
didn't have that requirement, if memory serves. OTOH, I'm not sure
how conforming code could see the difference, so it might still be
able to sneak in under the as-if rule.

 

[Frederick Gotham]

Jerry Coffin posted:

I should have pointed out that this was a C (89/90) compiler, which
didn't have that requirement, if memory serves. OTOH, I'm not sure
how conforming code could see the difference, so it might still be
able to sneak in under the as-if rule.

In C++, only the following three types are guaranteed to contain no
padding:

char
signed char
unsigned char


In C, only the following type is guaranteed to contain no padding:

unsigned char