Trap representations for unsigned integers

[Army1987]

If uMyInt_t is an unsigned integral type, is the following a
necessary and sufficient condition that uMyInt_t has no trap
representation?

(uMyInt_t)(-1) >> CHAR_BIT*sizeof(uMyInt_t)-1

That is, I'm asking wheter it equals 0 whenever uMyInt_t has trap
representations, equals a nonzero value whenever uMyInt_t has no
trap representation, and never triggers undefined behaviour.

 

[Guest]

If uMyInt_t is an unsigned integral type, is the following a
necessary and sufficient condition that uMyInt_t has no trap
representation?

(uMyInt_t)(-1) >> CHAR_BIT*sizeof(uMyInt_t)-1

No. If uMyInt_t has padding bits, you will right-shift by a number
greater than (or equal to) the number of value bits, and for that the
behaviour is undefined.

 

[Eric Sosman]

If uMyInt_t is an unsigned integral type, is the following a
necessary and sufficient condition that uMyInt_t has no trap
representation?

(uMyInt_t)(-1) >> CHAR_BIT*sizeof(uMyInt_t)-1

That is, I'm asking wheter it equals 0 whenever uMyInt_t has trap
representations, equals a nonzero value whenever uMyInt_t has no
trap representation, and never triggers undefined behaviour.

I think there are at least two problems with this test.
First, if uMyInt_t has padding bits the shift count may be
too large and lead to undefined behavior ("may" because of
possible promotion to int or unsigned int). Second, the
presence of padding bits does not imply the existence of trap
representations: the extra bits may just be along for the ride.

The best way to detect padding bits may be to count the
number of 1's in (uMyInt_t)-1, or to compare the numeric value
of (uMyInt_t)-1 to the "expected" quantity. The first test is
easy at run time but difficult or impossible at preprocessing
time; the second has problems, too (what type should you use
to form the expected value?). I can think of no reliable way
to determine whether trap representations exist; if you find
there are no padding bits you can deduce that there are no
traps, but that's as far as I think you can get.

 

[Army1987]

No. If uMyInt_t has padding bits, you will right-shift by a number
greater than (or equal to) the number of value bits, and for that the
behaviour is undefined.

Is

( DBL_MAX >= (uMyInt_t)(-1) || (puts("Your DS9K is about to self-\
destruct. Get a real computer."), exit(1)),
/* On the DeathStation 9000 exit(1) activates self-destruction */
ceil(log2((uMyInt_t)(-1))) >= CHAR_BIT*sizeof(uMyInt_t) )

any better?

 

[Guest]

Is

( DBL_MAX >= (uMyInt_t)(-1) || (puts("Your DS9K is about to self-\
destruct. Get a real computer."), exit(1)),
/* On the DeathStation 9000 exit(1) activates self-destruction */
ceil(log2((uMyInt_t)(-1))) >= CHAR_BIT*sizeof(uMyInt_t) )

any better?

On the DS9K, DBL_MAX would be large enough, but CHAR_BIT *
sizeof(uMuInt_t) would give the wrong result because SIZE_MAX is too
small.

Seriously though, your current expression is already no longer an
integer constant expression, so at this point there's no downside to
just writing a function.

 

[Army1987]

[snip] Second, the

presence of padding bits does not imply the existence of trap
representations: the extra bits may just be along for the ride.

So I'll replace "necessary and sufficient condition" with
"sufficient condition".

What I was thinking of is something like:

#include <string.h>
unsigned char randchar();
/* Get a random integer from 0 to UCHAR_MAX */

unsigned long longrand()
{
unsigned long result;
if (NO_TRAP(unsigned long)) {
int i;
unsigned char res[sizeof result];
for (i=0; i<sizeof result; i++)
res = randchar();
memcpy(&result, res, sizeof result);
} else {
/* invent something else */
}
return result;
}

I would still be able to use the algorithm with the right result (a
uniformly distributed random integer from 0 to UINT_MAX) if there
are padding bits but they are ignored.

 

[Army1987]

Army1987 wrote:

On the DS9K, DBL_MAX would be large enough, but CHAR_BIT *
sizeof(uMuInt_t) would give the wrong result because SIZE_MAX is too
small.

On C99 I might use (uintmax_t)CHAR_BIT*sizeof(uMyInt_t)...

Seriously though, your current expression is already no longer an
integer constant expression, so at this point there's no downside to
just writing a function.

I was thinking of using a macro, so I could write
NO_PADDING(size_t), NO_PADDING(unsigned int), etc. There's no way
to do that with a function.

 

[Guest]

[snip] Second, the

presence of padding bits does not imply the existence of trap
representations: the extra bits may just be along for the ride.

So I'll replace "necessary and sufficient condition" with
"sufficient condition".

What I was thinking of is something like:

#include <string.h>
unsigned char randchar();
/* Get a random integer from 0 to UCHAR_MAX */

unsigned long longrand()
{
unsigned long result;
if (NO_TRAP(unsigned long)) {
int i;
unsigned char res[sizeof result];
for (i=0; i<sizeof result; i++)
res = randchar();
memcpy(&result, res, sizeof result);
} else {
/* invent something else */
}
return result;
}

I would still be able to use the algorithm with the right result (a
uniformly distributed random integer from 0 to UINT_MAX) if there
are padding bits but they are ignored.

Why not

unsigned long result = randchar();
if ((unsigned long) -1 > UCHAR_MAX)
{
size_t i;
for (i = sizeof result - 1; i != 0; i--)
result = (result << CHAR_BIT) | randchar();
}

It works regardless of any padding bits.

 

[Guest]

On C99 I might use (uintmax_t)CHAR_BIT*sizeof(uMyInt_t)...

And on C90, you can use a cast to unsigned long.

I was thinking of using a macro, so I could write
NO_PADDING(size_t), NO_PADDING(unsigned int), etc. There's no way
to do that with a function.

You could do

#define NO_PADDING(type) (count_bits((type) -1) == sizeof(type))

where count_bits accepts an unsigned long / uintmax_t.

 

[Guest]

You could do

#define NO_PADDING(type) (count_bits((type) -1) == sizeof(type))

.... == sizeof(type) * CHAR_BIT

 

[Army1987]

Why not

unsigned long result = randchar();
if ((unsigned long) -1 > UCHAR_MAX)
{
size_t i;
for (i = sizeof result - 1; i != 0; i--)
result = (result << CHAR_BIT) | randchar();
}

It works regardless of any padding bits.

I hadn't thought of that before... (Mathematically x << y << z << t
is equivalent to x << (y+z+t), but I hadn't thought to the fact
that the former might be valid C even when the latter may be UB.)
Thanks.

 

[Peter Nilsson]

unsigned long result = randchar();
if ((unsigned long) -1 > UCHAR_MAX)
{
    size_t i;
    for (i = sizeof result - 1; i != 0; i--)
        result = (result << CHAR_BIT) | randchar();
}

It works regardless of any padding bits.

True, but it may call randchar() more times than needed
on a hypothetical implementation where there are LOTS
of padding bits.

unsigned long m = -1, result = randchar();
while ((m = m >> (CHAR_BIT - 1) >> 1) != 0)
result = (result << CHAR_BIT) | randchar();

 

[Army1987]

True, but it may call randchar() more times than needed
on a hypothetical implementation where there are LOTS
of padding bits.

unsigned long m = -1, result = randchar();
while ((m = m >> (CHAR_BIT - 1) >> 1) != 0)
result = (result << CHAR_BIT) | randchar();

Or something more readable:

#include <limits.h>
#if __STDC_VERSION__ >= 199901L
#include <stdint.h>
#else
#define uintmax_t unsigned long
#endif

int count_bits(uintmax_t n)
{
int result = 1;
while (n/=2)
result++;
return result;
}

#define PADDING(t) ( CHAR_BIT*sizeof(t) - count_bits((t)(-1)) )

unsigned long longrand()
{
size_t i;
unsigned long result = 0;
size_t bytes = sizeof result - PADDING(unsigned long)/CHAR_BIT;
for (i=0; i<bytes; i++) {
result <<= CHAR_BIT;
result |= randchar();
}
return result;
}

(Did I get this right?)

 

[Keith Thompson]

#include <limits.h>
#if __STDC_VERSION__ >= 199901L
#include <stdint.h>
#else
#define uintmax_t unsigned long
#endif

Why are you using #define rather than typedef?

The test for __STDC_VERSION__ *should* work, but unfortunately in real
life it may not be reliable. The __STDC_VERSION__ macro typically
tells you whether the compiler claims to support C99; this may or may
not tell you whether the <stdint.h> header exists. You can also get a
false negative, if the compiler doesn't claim to support C99, but it
supports unsigned long long as an extension. (Most such compilers
support a mode in which they conform to C90 without supporting
C99-specific features; the trick is to make sure the compiler is
actually invoked in such a mode.)

A conditional #include would be handy here:

#if header_exists <stdint.h>
#include <stdint.h>
#else
typedef unsigned long uintmax_t
#endif

but, alas, that doesn't exist.

Configuration systems like GNU autoconf can be helpful, but the
details are off-topic.

 

[pete]

True, but it may call randchar() more times than needed
on a hypothetical implementation where there are LOTS
of padding bits.

unsigned long m = -1, result = randchar();
while ((m = m >> (CHAR_BIT - 1) >> 1) != 0)
result = (result << CHAR_BIT) | randchar();

(result << CHAR_BIT) is undefined if sizeof(result) equals one.

 

[Army1987]

(result << CHAR_BIT) is undefined if sizeof(result) equals one.

if sizeof result is 1, then m is UCHAR_MAX, and
m >> (CHAR_BIT - 1) >> 1 is 0.
So the body loop is never done then.
(I still wonder why uns_var >> value_bits and uns_var << value_bits
are undefined. uns_var * 2**value_bits modulo 2**value_bits is 0,
and so is floor(uns_var / 2**value_bits). That's what
uns_var >> value_bits-1 >> 1 and uns_var << value_bits-1 << 1 do.
Maybe I'll post that to comp.std.c.)

 

[Flash Gordon]

Army1987 wrote, On 28/04/07 10:28:

(I still wonder why uns_var >> value_bits and uns_var << value_bits
are undefined. uns_var * 2**value_bits modulo 2**value_bits is 0,
and so is floor(uns_var / 2**value_bits). That's what
uns_var >> value_bits-1 >> 1 and uns_var << value_bits-1 << 1 do.
Maybe I'll post that to comp.std.c.)

The most probable reason for it being undefined is because the way
processors will treat it if you use assembler instructions to do it varies.

 

[Peter Nilsson]

Or something more readable:

But less correct. ;-)

#include <limits.h>

#define PADDING(t) ( CHAR_BIT*sizeof(t) - count_bits((t)(-1)))

Why are people so hung up on calculating the number of
padding bits?

unsigned long longrand()
{
size_t i;
unsigned long result = 0;
size_t bytes = sizeof result - PADDING(unsigned long)
/CHAR_BIT;
for (i=0; i<bytes; i++) {
result <<= CHAR_BIT;
result |= randchar();
}
return result;

}

(Did I get this right?)

No. If unsigned long is exactly 1 byte (ergo unpadded),
then the left shifting of result by CHAR_BIT bits invokes
undefined behaviour.