# Shift Operation

Discussion in 'C Programming' started by Nishu, Oct 9, 2006.

1. ### NishuGuest

Hi All,

Could you please explain whether C standard supports logical right
shift operation using some operator?

I know somewhere I read about >>> operator. I thought, it is in C but i
think i'm wrong about it. No where google helps me determining this
lapse in my memory. MSVC 6 compiler gives me error.

Thanks && Regards,
Nishu

Nishu, Oct 9, 2006

2. ### Richard HeathfieldGuest

Nishu said:

> Hi All,
>
> Could you please explain whether C standard supports logical right
> shift operation using some operator?

C's right shift operator, >>, works like this:

A >> B will yield a result that is equivalent to A shifted right through B
bit positions. If A is negative, the result is implementation-defined.

There is also a >>= operator, of course, so that you can do A >>= B instead
of A = A >> B

> I know somewhere I read about >>> operator. I thought, it is in C

No, 'fraid not.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

Richard Heathfield, Oct 9, 2006

3. ### NishuGuest

Richard Heathfield wrote:
> Nishu said:
>
> > Hi All,
> >
> > Could you please explain whether C standard supports logical right
> > shift operation using some operator?

>
> C's right shift operator, >>, works like this:
>
> A >> B will yield a result that is equivalent to A shifted right through B
> bit positions. If A is negative, the result is implementation-defined.

If i take a signed integer, so right shifting a negative int should
give me another negative int (signed arithmetic) or a positive integer.
Is this C standard defines or implementation defines?

Actually i want to know whether
long A;
A = 0xFFFFFFFF;

A >>= 1;

A &= 0x80000000

if(A)
{
printf("operator is arithmetic right shift");
}
else
{
printf(" operator is logical right shift");
}

Thanks.
Nishu

Nishu, Oct 9, 2006
4. ### Richard BosGuest

"Nishu" <> wrote:

> Could you please explain whether C standard supports logical right
> shift operation using some operator?

No. Or rather, not reliably.

Right shifting on unsigned integers is, of course, neither logical nor
arithmetical (or both, if you wish); zeros get shifted in from the high
end, but there is no sign bit to copy or not to copy.

Right shifting on signed integers is different for positive and negative
signed integers.
If the signed integer has a positive or zero value, zero bits are
shifted in from the high end just as for unsigned types; it makes no
difference whether this is because it is a logical shift, or whether
it's an arithmetical shift with a zero sign bit being copied. A zero bit
is, after all, equal to any other zero bit.
If, however, the signed integer has a negative value, the resulting
value is implementation-defined. This means that your program is not
allowed to crash on this operation[1], but the Standard doesn't require
any particular result. All it requires is that your implementation
defines what the result will be. This could be a logical shift, an
arithmetical shift, or something entirely different (which might makes
sense on, e.g., one's-complement machines).

So in short, _your compiler_ might use a logical right shift if you use
the normal C >> shifting operator, but that assumption is not portable:
there's no guarantee that this will work on the next platform you try it
on.

> I know somewhere I read about >>> operator. I thought, it is in C but i
> think i'm wrong about it.

There is indeed no such thing.

Richard

[1] Unlike _left_-shifting a signed integer where the result would
overflow; that is undefined, and may therefore crash.

Richard Bos, Oct 9, 2006
5. ### Walter RobersonGuest

In article <>,
Nishu <> wrote:

>If i take a signed integer, so right shifting a negative int should
>give me another negative int (signed arithmetic) or a positive integer.
>Is this C standard defines or implementation defines?

Implementation. Except that those aren't the only two choices
available to the implementation...

>Actually i want to know whether

>long A;
>A = 0xFFFFFFFF;

>A >>= 1;

>A &= 0x80000000

>if(A)
>{
>printf("operator is arithmetic right shift");
>}
>else
>{
>printf(" operator is logical right shift");
>}

You are presuming that A = 0xFFFFFFFF will store a negative number
in A. That is a bad assumption:

1) long might have more than 32 bits, in which case 0xFFFFFFFF
would just be a regular signed number. It is not uncommon for long
to be 64 bits with int being 32 bits, short 16 bits, char 8 bits.
But it is also not uncommon for int and long both to be the same size
of 32 bits; there are also a number of compilers for which
long is 32 bits, int is 16 bits...

2) 0xFFFFFFFF is not specifically indicated as a long constant via an 'L'
suffix; interpretation of it starts out by considering it as a
signed int. No negative sign is present in the number, so the
compiler will inspect to determine whether 0xFFFFFFFF fits within
the positive range of signed int on that system; if it does then
0xFFFFFFFF would be considered a positive signed int and there would
then be an implicit cast of that positive signed int into a long for
storage into A; as long is promised to be at least as wide as int,
that would either involve leaving the number unchanged or else
widening it if necessary; widening on most systems would involve
sticking the value in the low bits and zero-filling the upper bits,
but int and long need not have the same internal padding bit structures
so an actual representation change might take place.

If the compiler determines that 0xFFFFFFFF does not fit within
the positive range of signed int, then it would reconsider it as
potentialy being a positive signed long; if it does not fit within
a positive signed long, then it would convert it to unsigned long, which
it should fit into. So you would now have an unsigned long constant
token and you would have a simple long destination to store it into.
The C standard says that if you attempt to store an unsigned
value into a signed location, and the unsigned value fits within
the positive range of the signed type, then the positive value
will be stored -- but it also says that if the unsigned value does
*not* fit within the positive range of the signed type, that the
result of the conversion is up to the implementation. Thus,
long A = 0xFFFFFFFF is not necessarily going to produce a negative
result in A, even if the implementation happens to use 32 bit long.
--
I was very young in those days, but I was also rather dim.
-- Christopher Priest

Walter Roberson, Oct 9, 2006
6. ### Richard HeathfieldGuest

Nishu said:

> Richard Heathfield wrote:

<snip>

>> If A is negative, the result is implementation-defined.

>
> If i take a signed integer, so right shifting a negative int should
> give me another negative int (signed arithmetic) or a positive integer.
> Is this C standard defines or implementation defines?

If A is negative, the result is implementation-defined.

> Actually i want to know whether
> long A;
> A = 0xFFFFFFFF;
>
> A >>= 1;
>
> A &= 0x80000000
>
> if(A)
> {
> printf("operator is arithmetic right shift");
> }
> else
> {
> printf(" operator is logical right shift");
> }

If A is negative, the result is implementation-defined.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

Richard Heathfield, Oct 9, 2006
7. ### NishuGuest

Walter Roberson wrote:
> In article <>,
> Nishu <> wrote:
>
> >If i take a signed integer, so right shifting a negative int should
> >give me another negative int (signed arithmetic) or a positive integer.
> >Is this C standard defines or implementation defines?

>
> Implementation. Except that those aren't the only two choices
> available to the implementation...

< snip>

> The C standard says that if you attempt to store an unsigned
> value into a signed location, and the unsigned value fits within
> the positive range of the signed type, then the positive value
> will be stored -- but it also says that if the unsigned value does
> *not* fit within the positive range of the signed type, that the
> result of the conversion is up to the implementation. Thus,
> long A = 0xFFFFFFFF is not necessarily going to produce a negative
> result in A, even if the implementation happens to use 32 bit long.

Thanks. I got it. C is indeed very flexible and benevolent to
compilers.

-Nishu

Nishu, Oct 9, 2006
8. ### Guest

Nishu wrote:

> I know somewhere I read about >>> operator. I thought, it is in C but i
> think i'm wrong about it.

>>> (unsigned right shift) exists in Java <http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.19>, but not in standard C.

, Oct 9, 2006
9. ### Frederick GothamGuest

Nishu posted:

> I know somewhere I read about >>> operator.

Maybe you could write one yourself? If you know that "unsigned int" and
"signed int" have no padding, then you could simply do:

int i = -57;

*(unsigned*)&i >>= 4;

Or maybe something like:

i < 0 ? (i = -i) >>= 4, i = -i : i >>= 4

--

Frederick Gotham

Frederick Gotham, Oct 9, 2006
10. ### Flash GordonGuest

Frederick Gotham wrote:
> Nishu posted:
>
>> I know somewhere I read about >>> operator.

Where >>> is meant to be a logical shift right.

> Maybe you could write one yourself? If you know that "unsigned int" and
> "signed int" have no padding, then you could simply do:
>
> int i = -57;
>
> *(unsigned*)&i >>= 4;

Given your limitations I believe this would work.

> Or maybe something like:
>
> i < 0 ? (i = -i) >>= 4, i = -i : i >>= 4

This one will overflow with INT_MIN if INT_MIN == -INT_MAX - 1, i.e. on
most 2s complement machines.
--
Flash Gordon

Flash Gordon, Oct 9, 2006
11. ### Peter NilssonGuest

Frederick Gotham wrote:
> Nishu posted:
> > I know somewhere I read about >>> operator.

>
> Maybe you could write one yourself? If you know that "unsigned int" and
> "signed int" have no padding, then you could simply do:
>
> int i = -57;
>
> *(unsigned*)&i >>= 4;

This is legal in C99 [not sure about C90]. But you may not get the
'desired' result for sm or 1c machines.

--
Peter

Peter Nilsson, Oct 10, 2006
12. ### NishuGuest

Frederick Gotham wrote:
> Nishu posted:
>
> > I know somewhere I read about >>> operator.

>
>
> Maybe you could write one yourself? If you know that "unsigned int" and
> "signed int" have no padding, then you could simply do:
>
> int i = -57;
>
> *(unsigned*)&i >>= 4;

Typecasting is certainly good option. Thanks.

(unsigned) i >>= 1; /* (Results in logical shift) */

What is the purpose of doing it like *(unsigned*)&i >>= 1; ?

-Nishu

Nishu, Oct 10, 2006
13. ### Chris DollinGuest

Nishu wrote:

> Frederick Gotham wrote:
>> Nishu posted:
>>
>> > I know somewhere I read about >>> operator.

>>
>> Maybe you could write one yourself? If you know that "unsigned int" and
>> "signed int" have no padding, then you could simply do:
>>
>> int i = -57;
>>
>> *(unsigned*)&i >>= 4;

>
> Typecasting is certainly good option. Thanks.
>
> (unsigned) i >>= 1; /* (Results in logical shift) */

Results in a disgnostic message, I hope. Cast-expressions are not
lvalues.

> What is the purpose of doing it like *(unsigned*)&i >>= 1; ?

To bypass (in a somewhat clunky way) the restriction than cast-expressions
are not lvalues. I /think/ that you still get undefined behaviour, but
it might only be implementation-defined. Or I might be wrong. We're
having a warm October; maybe Hell is leaking.

--
Chris "Essen -9 and counting" Dollin
"I'm still here and I'm holding the answers" - Karnataka, /Love and Affection/

Chris Dollin, Oct 10, 2006
14. ### Richard TobinGuest

In article <qDxWg.14659\$>,
Frederick Gotham <> wrote:

> *(unsigned*)&i >>= 4;

Why not use the more transparent

i = ((unsigned)i) >> 4;

Taking the address of a variable may well prevent the compiler from
putting it in a register, though a sufficiently clever compiler would
not be fooled.

-- Richard

Richard Tobin, Oct 10, 2006
15. ### Ancient_HackerGuest

Nishu wrote:

> If i take a signed integer, so right shifting a negative int should
> give me another negative int (signed arithmetic)

Why would you ever want to do an "arithmetic" right shift on a negative
number?

Even if the "sign" bit got shifted into the top bits, the result is
mostly mathematically useless. i.e. shift of -1 on a two's complement
machine gives you .....

Ancient_Hacker, Oct 10, 2006
16. ### Richard TobinGuest

In article <>,
Ancient_Hacker <> wrote:

>Why would you ever want to do an "arithmetic" right shift on a negative
>number?

To divide by 2?

>Even if the "sign" bit got shifted into the top bits, the result is
>mostly mathematically useless. i.e. shift of -1 on a two's complement
>machine gives you .....

-1, which is (-1)/2 rounded down (which is the behaviour I usually
want from integer division).

-- Richard

Richard Tobin, Oct 10, 2006
17. ### Frederick GothamGuest

Nishu posted:

> (unsigned) i >>= 1; /* (Results in logical shift) */

As Chris mentioned, you might want to set your compiler to Strict C Mode if
it compiles that for you. No case ever results in an L-value in C.

--

Frederick Gotham

Frederick Gotham, Oct 10, 2006
18. ### Frederick GothamGuest

Richard Tobin posted:

> i = ((unsigned)i) >> 4;

If I'm not mistake, this will not retain the original bit-pattern on systems
other than 2's complement.

--

Frederick Gotham

Frederick Gotham, Oct 10, 2006
19. ### Frederick GothamGuest

Frederick Gotham posted:

> No case ever results in an L-value in C.

Typo correction:

No CAST ever results in a...

--

Frederick Gotham

Frederick Gotham, Oct 10, 2006
20. ### =?utf-8?B?SGFyYWxkIHZhbiBExLNr?=Guest

Frederick Gotham wrote:
> Richard Tobin posted:
> > i = ((unsigned)i) >> 4;

>
> If I'm not mistake, this will not retain the original bit-pattern on systems
> other than 2's complement.

If you care about the bit pattern, you probably shouldn't be using
signed integers anyway.

=?utf-8?B?SGFyYWxkIHZhbiBExLNr?=, Oct 10, 2006