what does it do?

[puzzlecracker]

what does it do?
#define BLACKBOX(x) ((x)&((x)-1))

[Michael Mair]

puzzlecracker wrote:

what does it do?
#define BLACKBOX(x) ((x)&((x)-1))

For unsigned integers x and signed integers x in 2s complement
with x>(integer type's minimum value), this clears the least
significant set bit.
It also works for 1s complement signed integers and
x!=0 && x!=1 && x>(integer type's minimum value)

-Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[Luke Wu]

puzzlecracker wrote:

what does it do?
#define BLACKBOX(x) ((x)&((x)-1))

Clears the least significant set bit in x. This works on all unsigned
numbers and all signed numbers (except "sign-magnitude" negative even
numbers).

[Michael Mair]

Luke Wu wrote:

puzzlecracker wrote:

what does it do?
#define BLACKBOX(x) ((x)&((x)-1))

Clears the least significant set bit in x. This works on all unsigned
numbers and all signed numbers (except "sign-magnitude" negative even
numbers).

.... and except for x==T_MIN where T is the respective integer type;
in the case of 1s complement you cannot rely on 0 not switching
its representation, so x==0 and x==1 are not guaranteed to work
either.
Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[Peter Nilsson]

> puzzlecracker wrote:

what does it do?
#define BLACKBOX(x) ((x)&((x)-1))

What do you think it does?

If you want to 'test' clc, at least try to be original.

Michael Mair wrote:
For unsigned integers x and signed integers x in 2s complement
with x>(integer type's minimum value), this clears the least
significant set bit.
It also works for 1s complement signed integers and
x!=0 && x!=1 && x>(integer type's minimum value)

No, consider the following (high bit is sign-bit):

1c or 2c(C99) sm
x : 10000000 10000000 10000000 01111111
x - 1: 10000000 01111111 10000000 10000000
& : 10000000 00000000 10000000 00000000

The result could be a trap representation.

Note also that C89 potentially allows negative representations
beyond 2c, 1c and sm.

--
Peter

[Raymond Martineau]

On 21 Feb 2005 13:08:16 -0800, "Luke Wu" <Lo***********@gmail.com> wrote:

puzzlecracker wrote:

what does it do?
#define BLACKBOX(x) ((x)&((x)-1))

Clears the least significant set bit in x. This works on all unsigned
numbers and all signed numbers (except "sign-magnitude" negative even
numbers).

Inputting the integer '8' causes the "function" to return zero. Not
suprizing, since the original poster is a known troll.

[Ben Pfaff]

bk***@ncf.ca (Raymond Martineau) writes:

On 21 Feb 2005 13:08:16 -0800, "Luke Wu" <Lo***********@gmail.com> wrote:

puzzlecracker wrote:

what does it do?
#define BLACKBOX(x) ((x)&((x)-1))

Clears the least significant set bit in x. This works on all unsigned
numbers and all signed numbers (except "sign-magnitude" negative even
numbers).

Inputting the integer '8' causes the "function" to return zero. Not
suprizing, since the original poster is a known troll.

8 has only one bit set. Clearing that bit, which is necessarily
the least significant set bit, yields 0. Thus, this is correct
behavior.
--
int main(void){char p[]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv wxyz.\
\n",*q="kl BIcNBFr.NKEzjwCIxNJC";int i=sizeof p/2;char *strchr();int putchar(\
);while(*q){i+=strchr(p,*q++)-p;if(i>=(int)sizeof p)i-=sizeof p-1;putchar(p[i]\
);}return 0;}

[jjf]

Raymond Martineau wrote:

On 21 Feb 2005 13:08:16 -0800, "Luke Wu" <Lo***********@gmail.com> wrote:

puzzlecracker wrote:

what does it do?
#define BLACKBOX(x) ((x)&((x)-1))
Clears the least significant set bit in x. This works on all unsignednumbers and all signed numbers (except "sign-magnitude" negative evennumbers).

Inputting the integer '8' causes the "function" to return zero.

Yes, exactly as Luke said. What's your point?
Not suprizing, since the original poster is a known troll.

What are you on about?

[Michael Mair]

Peter Nilsson wrote:

puzzlecracker wrote:

what does it do?
#define BLACKBOX(x) ((x)&((x)-1))

What do you think it does?

If you want to 'test' clc, at least try to be original.

Ack.
Michael Mair wrote:

For unsigned integers x and signed integers x in 2s complement
with x>(integer type's minimum value), this clears the least
significant set bit.
It also works for 1s complement signed integers and
x!=0 && x!=1 && x>(integer type's minimum value)
No, consider the following (high bit is sign-bit):

1c or 2c(C99) sm
x : 10000000 10000000 10000000 01111111
x - 1: 10000000 01111111 10000000 10000000
& : 10000000 00000000 10000000 00000000

The result could be a trap representation.

Just to make sure: Your "no" is referring to the fact that we
could get a trap representation? Or do you mean that the
formulation "least significant bit" is incorrect?
I am not sure what you want to say with the sign-magnitude example
-- most certainly the "lowest" bit of x is not cleared.

Note also that C89 potentially allows negative representations
beyond 2c, 1c and sm.

Thank you for the reminder (mostly, I assume C99 _restrictions_
for everything and am surprised if C89 is not as strict).
Cheers
Michael
--
E-Mail: Mine is a gmx dot de address.