Byte parity question...

[philbo30]

I admit, I didn't spend any time researching this yet so this is the
first step...

Is there a C function available to determine byte parity or is the
usual course to just count the 1s?

[cr88192]

"philbo30" <ma******@gmail.comwrote in message
news:11**********************@22g2000hsm.googlegro ups.com...

>I admit, I didn't spend any time researching this yet so this is the
first step...

Is there a C function available to determine byte parity or is the
usual course to just count the 1s?

counting 1s, though possible, but is a slow option.

k=0; for(j=0; j<8; j++)if(i&(1<<j))k++;
even: !(k&1)
odd: k&1
one can use a table, or a bitmap, instead.

nibbles (Even, Odd):
0-E 1-O 2-O 3-E
4-O 5-E 6-E 7-O
8-O 9-E A-E B-O
C-E D-O E-O F-E

so, a bitmask:
static int p_even=0x9669;
static int p_odd=0x6996;

nibble 'i' is even? 'p_even&(1<<i)' or '(p_even>>i)&1'.

and for a byte:
((p_even>>(i&15))&1)==((p_even>>((i>>4)&15))&1)

or, odd:
((p_odd>>(i&15))^(p_odd>>((i>>4)&15)))&1

consequently, we can define a bitmap table, and do a byte like this:
(p_even_bm[i>>4]>>(i&15))&1
(p_odd_bm[i>>4]>>(i&15))&1
or, even faster (but taking a lot more space), is giving an individual value
for every input value:
p_even_tab[i]
p_odd_tab[i]

along with other possibilities...

[junky_fellow]

On Sep 10, 4:13 am, philbo30 <[e-mail]:

I admit, I didn't spend any time researching this yet so this is the
first step...

Is there a C function available to determine byte parity or is the
usual course to just count the 1s?

You may use XOR operation.

[Charlie Gordon]

<ju**********@yahoo.co.ina écrit dans le message de news:
11*********************@50g2000hsm.googlegroups.co m...

On Sep 10, 4:13 am, philbo30 <[e-mail]:

>I admit, I didn't spend any time researching this yet so this is the
first step...

Is there a C function available to determine byte parity or is the
usual course to just count the 1s?

You may use XOR operation.

That's an interesting alternative to a memory based lookup table.
Assuming bytes have 8 bits, this code should do:

static inline int byte_parity(unsigned char b) {
b ^= b >4;
b ^= b >2;
b ^= b >1;
return b & 1;
}

This method can be easily generalized to larger unsigned types.

--
Chqrlie

[Tor Rustad]

philbo30 wrote:

I admit, I didn't spend any time researching this yet so this is the
first step...

Is there a C function available to determine byte parity or is the
usual course to just count the 1s?

There is no such function in the standard C library.

If CHAR_BIT is 8, you could use this table:

static unsigned char parity[] =
{
0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,0,
1,1,0,1,0,0,1,1,0,0,1,0,1,1,0,0,1,1,0,1,0,0,1,0,1,
1,0,1,0,0,1,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,0,1,1,
0,1,0,0,1,0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0,0,1,1,0,
1,0,0,1,1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,0,1,1,0,1,
0,0,1,1,0,0,1,0,1,1,0,0,1,1,0,1,0,0,1,0,1,1,0,1,0,
0,1,1,0,0,1,0,1,1,0,0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,
0,1,0,0,1,0,1,1,0,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,
1,0,0,1,0,1,1,0,1,0,0,1,0,1,1,0,0,1,1,0,1,0,0,1,1,
0,0,1,0,1,1,0,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,1,0,
0,1,0,1,1,0
};

where parity[octet] gives the bit count. Yes, the above table was
generated by simply counting the 1s.

--
Tor <torust [at] online [dot] no>

[Tor Rustad]

Tor Rustad wrote:

where parity[octet] gives the bit count.

ITYM:

"where parity[octet] gives the parity of the octet."
--
Tor <torust [at] online [dot] no>

[Richard Tobin]

In article <H9*********************@telenor.com>,
Tor Rustad <to********@hotmail.comwrote:

>where parity[octet] gives the bit count.

>ITYM:

"where parity[octet] gives the parity of the octet."

Maybe he just forgot to say "modulo 2" at the end.

-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.

[Thad Smith]

Charlie Gordon wrote:

<ju**********@yahoo.co.ina écrit dans le message de news:
11*********************@50g2000hsm.googlegroups.co m...

>On Sep 10, 4:13 am, philbo30 <[e-mail]:

>>Is there a C function available to determine byte parity or is the
usual course to just count the 1s?

Assuming bytes have 8 bits, this code should do:

static inline int byte_parity(unsigned char b) {
b ^= b >4;
b ^= b >2;
b ^= b >1;
return b & 1;
}

This method can be easily generalized to larger unsigned types.

This generalization, for a byte, should work for all sizes of bytes.
Most compilers will eliminate the while loop for 8-bit byte targets.

/* Function: parity
** Description: Return the parity of the argument.
*/
int /* parity of b: 1 if number of 1 bits is odd,
** 0 otherwise. */
parity (unsigned char b) {
while (b 0xff) {
b = (b >8) ^ (b & 0xff);
}
b ^= b >4;
b ^= b >2;
return ((b >1) ^ b) & 1;
}

--
Thad

[Michal Nazarewicz]

>On Sep 10, 4:13 am, philbo30 <[e-mail]:

>>Is there a C function available to determine byte parity or is the
usual course to just count the 1s?

Charlie Gordon wrote:

>Assuming bytes have 8 bits, this code should do:

static inline int byte_parity(unsigned char b) {
b ^= b >4;
b ^= b >2;
b ^= b >1;
return b & 1;
}

This method can be easily generalized to larger unsigned types.

Thad Smith <Th*******@acm.orgwrites:

This generalization, for a byte, should work for all sizes of
bytes. Most compilers will eliminate the while loop for 8-bit byte
targets.

Or one could use preprocessor directives to help compiler a bit:

/* Function: parity
** Description: Return the parity of the argument.
*/
int /* parity of b: 1 if number of 1 bits is odd,
** 0 otherwise. */
parity (unsigned char b) {

#if CHAR_BIT 16

while (b 0xff) {
b = (b >8) ^ (b & 0xff);
}

#elif CHAR_BIT 8
b ^= b >8;
#endif

b ^= b >4;
b ^= b >2;
return ((b >1) ^ b) & 1;
}

--
Best regards, _ _
.o. | Liege of Serenly Enlightened Majesty of o' \,=./ `o
..o | Computer Science, Michal "mina86" Nazarewicz (o o)
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