bit shifting

[digi.empire]

trying to wrap my mind around these two exercises:

1. given an int (which is 16 bits in c), if the 16 bits are:

A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0

it will output an int in whose 16 bits are:

A7A6A5A4A3A2A1A0A15A14A13A12A11A10A9A8

2. given an int(which is 16bits in c) if the 16 bits are:

A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0

it will perform a cirular right shift 5 so that the new int's 16 bits
are:

A12A11A10A9A8A7A6A5A4A3A2A1A0A15A14A13

thanks in advanced.

[Jens Thoms Toerring]

di*********@gmail.com wrote:

trying to wrap my mind around these two exercises:

1. given an int (which is 16 bits in c),

No, in C an int has *at least* 16 bits, more are allowed (and
nowadays often the default).

if the 16 bits are:

A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0

That's not 16 bits in any notation I have seen.

it will output

What's "it"?

an int in whose 16 bits are:

A7A6A5A4A3A2A1A0A15A14A13A12A11A10A9A8

Again, that doesn't look like bits. Does it mean bit positions
(e.g. A7 standing for the 7th bit, if we start conuting at zero)?
In that case this looks like a rotation by 8 positions.

2. given an int(which is 16bits in c) if the 16 bits are:

A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0

it will perform a cirular right shift 5 so that the new int's 16 bits
are:

A12A11A10A9A8A7A6A5A4A3A2A1A0A15A14A13

What's your question? And how does it relate to C? Were you asked
to come up with a line of C that converts a (16-bit integer) from
the first to the second form? In that case read about the bitwise
OR operator ('|') plus the left and right shift operators, i.e.
'<<' and '>>' (but which can only be used with unsigned integers).
Here's a hint: what you need is going to look similar to

y = ( x >S ) | ( x << T );

(but there are also other ways to write the same and you may have
to logically and ('&') x with 0xFFFF before the shifts if an int
has more than 16 bits on your system) where x and y are (unsigned)
integers. All you still need to figure out are what values to use
for S and T for both the exercises. You can be pretty sure that
S + T will be 16;-)

If this is homework (and that's what it looks like) you better
post what you have tried so far and what your problem actually
is...
Regards, Jens
--
\ Jens Thoms Toerring __ _ [e-mail]
\__________________________ http://toerring.de

[Barry Schwarz]

On 13 Dec 2006 17:44:08 -0800, di*********@gmail.com wrote:

>trying to wrap my mind around these two exercises:

1. given an int (which is 16 bits in c), if the 16 bits are:

An int must be at least 16 bits but there are a lot of systems where
it is larger.

>
A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0

Since bits are always 1 or 0, what do the 38 characters above
represent?
Remove del for email

[Peter Nilsson]

Jens Thoms Toerring wrote:

... read about the bitwise OR operator ('|') plus the left and right
shift operators, i.e. '<<' and '>>' (but which can only be used
with unsigned integers).

s/can/should/

The bitwise operators can be used on signed integers too, albeit
with twice as many caveats. [Beware integral promotion when
operating on unsigned short and unsigned char.]

--
Peter

[David T. Ashley]

<di*********@gmail.comwrote in message
news:11*********************@79g2000cws.googlegrou ps.com...

trying to wrap my mind around these two exercises:

1. given an int (which is 16 bits in c), if the 16 bits are:

A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0

it will output an int in whose 16 bits are:

A7A6A5A4A3A2A1A0A15A14A13A12A11A10A9A8

(arg << 8) | ((arg >>8) & 0xFF) /* Mask necessary due to right shift
possible sign propagation. */

2. given an int(which is 16bits in c) if the 16 bits are:

A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0

it will perform a cirular right shift 5 so that the new int's 16 bits
are:

A12A11A10A9A8A7A6A5A4A3A2A1A0A15A14A13

(arg << 3) | (arg >13 & 0x7)

thanks in advanced.

Can the rest of us get a grade in the class, too?

[Neil]

Jens Thoms Toerring wrote:

di*********@gmail.com wrote:

>trying to wrap my mind around these two exercises:

>1. given an int (which is 16 bits in c),

No, in C an int has *at least* 16 bits, more are allowed (and
nowadays often the default).

>if the 16 bits are:

>A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0

That's not 16 bits in any notation I have seen.

>it will output

What's "it"?

> an int in whose 16 bits are:

>A7A6A5A4A3A2A1A0A15A14A13A12A11A10A9A8

Again, that doesn't look like bits. Does it mean bit positions
(e.g. A7 standing for the 7th bit, if we start conuting at zero)?
In that case this looks like a rotation by 8 positions.

>2. given an int(which is 16bits in c) if the 16 bits are:

>A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0

>it will perform a cirular right shift 5 so that the new int's 16 bits
are:

>A12A11A10A9A8A7A6A5A4A3A2A1A0A15A14A13

What's your question? And how does it relate to C? Were you asked
to come up with a line of C that converts a (16-bit integer) from
the first to the second form? In that case read about the bitwise
OR operator ('|') plus the left and right shift operators, i.e.
'<<' and '>>' (but which can only be used with unsigned integers).
Here's a hint: what you need is going to look similar to

y = ( x >S ) | ( x << T );

(but there are also other ways to write the same and you may have
to logically and ('&') x with 0xFFFF before the shifts if an int
has more than 16 bits on your system) where x and y are (unsigned)
integers. All you still need to figure out are what values to use
for S and T for both the exercises. You can be pretty sure that
S + T will be 16;-)

If this is homework (and that's what it looks like) you better
post what you have tried so far and what your problem actually
is...
Regards, Jens
--
\ Jens Thoms Toerring __ _ [e-mail]
\__________________________ http://toerring.de

If you stare at it long enough
A15,A14,A13,A12,A11,A10,A9,A8,A7,A6,A5,A4,A3,A2,A1 ,A0

[Sourcerer]

"Jens Thoms Toerring" <[e-mail] in message
news:4u*************@mid.uni-berlin.de...

di*********@gmail.com wrote:

<snip>

>A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0

That's not 16 bits in any notation I have seen.

Ax represents a bit. In the above, A15 is the highest value bit, and A0 is the
lowest value bit.

--
"It is easy in the world to live after the world's oppinion; it easy in solitude
to live after our own; but the great man is he who in the midst of the crowd
keeps with perfect sweetness the independence of solitude."
Ralph Waldo Emerson, Self-reliance 1841
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