Velocity Reviews > C++ > boolean logic question

# boolean logic question

Kurt Krueckeberg
Guest
Posts: n/a

 09-23-2004
In the book C++ Gothcas, Gotcha #7 is an example of using boolean logic to
simply code. My question follows this snippet from the book.

"Do you have to count to eight when presented with the following?"
int ctr = 0;
for (int i =0; i < 8; ++i) {
if (options & 1 << (8+i) )
if ( ctr++) {
cerr << "too many options selected";
break;
}

typedef unsigned short Bits;
inline Bits repeated( Bits b, Bits m)
{ return b & m & (b & m) -1; }
//. . .
if ( repeated (options, 0XFF) )
cerr << "Too many options slected";

My Question: Why can't repeated() be written simply as
inline Bits repeated (Bits b, Bits m)
{ return b & m;}

Why is the "& (b & m) - 1" necessary? What is that all about?

Thanks,
Kurt

Simon Stienen
Guest
Posts: n/a

 09-23-2004
Kurt Krueckeberg <(E-Mail Removed)> wrote:
> In the book C++ Gothcas, Gotcha #7 is an example of using boolean logic to
> simply code. My question follows this snippet from the book.
>
> "Do you have to count to eight when presented with the following?"
> int ctr = 0;
> for (int i =0; i < 8; ++i) {
> if (options & 1 << (8+i) )
> if ( ctr++) {
> cerr << "too many options selected";
> break;
> }
>
> typedef unsigned short Bits;
> inline Bits repeated( Bits b, Bits m)
> { return b & m & (b & m) -1; }
> //. . .
> if ( repeated (options, 0XFF) )
> cerr << "Too many options slected";
>
> My Question: Why can't repeated() be written simply as
> inline Bits repeated (Bits b, Bits m)
> { return b & m;}
>
> Why is the "& (b & m) - 1" necessary? What is that all about?
>
> Thanks,
> Kurt

With b & m you get a bitmask. If at least one bit is set, one bit will be
the highest set bit, for example: 0b00010000
If this is the only set bit then x-1 will be a mask with every bit up to
and including the highest set bit being 0 and every less significant bit
set. In this example: 0b00001111. Of course, a bitwise AND of those two
values will return 0.
On the other hand, if another bit was set, too (lets say 0b00010100), the
least significant set bit will be reset and all following bits are set,
resulting in the most significant bit staying set: 0b00010011.
Since the bit stays set, the bitwise AND won't reset this bit and the
result is non-zero.

HTH
Simon
--
Simon Stienen <http://dangerouscat.net> <http://slashlife.de>
»What you do in this world is a matter of no consequence,
The question is, what can you make people believe that you have done.«
-- Sherlock Holmes in "A Study in Scarlet" by Sir Arthur Conan Doyle

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are On Pingbacks are On Refbacks are Off Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post rickman VHDL 9 04-02-2011 05:17 AM Metre Meter Javascript 7 08-06-2010 08:40 PM J Leonard Java 4 01-19-2008 02:56 AM Miss. Michelle Heigardt Java 6 01-19-2004 09:08 AM Christopher Benson-Manica C Programming 7 12-30-2003 06:05 PM