lshift & rshift

[iesvs@free.fr]

Hello guys,
every time a rode a doc or a book about the language C I saw that
operators << and >exist. But each time they said that << translate
the digit to the left (and >...) but no one said if a << 1 mean
every time a * 2, because there is the problem with the way the
integer are stored. So I never use thoses operators because I fear
strange behaviour on different architecture. Can I use them and be
sure that a << 1 means a * 2 on every computer ? Is is in the ansi
standard of language C ?

[Richard Heathfield]

Re: lshift &amp; rshift

iesvs@free.fr said:

Hello guys,
every time a rode a doc or a book about the language C I saw that
operators << and >exist. But each time they said that << translate
the digit

Each bit is moved one place to the left, the leftmost bit falls off the end
into the bit bucket (which should be emptied regularly), and the rightmost
bit is set to 0.

to the left (and >...)

Beware signed types when shifting. The danger is particularly - er -
dangerous when right-shifting negative numbers. What happens to the sign
bit? (Answer: it depends on the implementation. )

but no one said if a << 1 mean
every time a * 2,

Not every time, no. For example, consider a system where CHAR_BIT is 8.

unsigned char ch = 1;

ch <<= 1; /* ch is now 2 */
ch <<= 1; /* ch is now 4 */
ch <<= 1; /* ch is now 8 */
ch <<= 1; /* ch is now 16 */
ch <<= 1; /* ch is now 32 */
ch <<= 1; /* ch is now 64 */
ch <<= 1; /* ch is now 128 */
ch <<= 1; /* ch is now 0, even though 2 * 128 is 256. */

because there is the problem with the way the
integer are stored. So I never use thoses operators because I fear
strange behaviour on different architecture. Can I use them and be
sure that a << 1 means a * 2 on every computer ?

If you stick to unsigned types, and don't push any 1-bits off the end, yes.
But why bother? Why not just multiply by 2, if that's what you want to do?

Is is in the ansi standard of language C ?

Yes, but - like I said - you do need to be careful.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Philip Potter]

Re: lshift &amp; rshift

iesvs@free.fr wrote:

Hello guys,
every time a rode a doc or a book about the language C I saw that
operators << and >exist. But each time they said that << translate
the digit to the left (and >...) but no one said if a << 1 mean
every time a * 2, because there is the problem with the way the
integer are stored. So I never use thoses operators because I fear
strange behaviour on different architecture. Can I use them and be
sure that a << 1 means a * 2 on every computer ? Is is in the ansi
standard of language C ?

One of the problems with using the bit-shift operators to do arithmetic is you
no longer say what you mean. If you want to multiply something by 2, say so.

Another problem is that bit-shifting is not entirely defined for signed types.
This is partly because a bit-shift could move a value bit into a sign bit, or
vice versa, and partly because C allows different representations of signed
integers (2's complement, 1's complement, or sign-magnitude).

In the bad old days, some people used to use bit-shifts to perform simple
multiplications because it would run quicker. These days, compilers can spot for
themselves when this sort of trick is possible and will do so.

For arithmetic, use arithmetic operators. Save bit-shifting for when you need
bit-shifting semantics - for example, simulating a shift register.

--
Philip Potter pgp <atdoc.ic.ac. uk

[iesvs@free.fr]

Re: lshift &amp; rshift

In the bad old days, some people used to use bit-shifts to perform simple

multiplications because it would run quicker. These days, compilers can spot for
themselves when this sort of trick is possible and will do so.
>
For arithmetic, use arithmetic operators. Save bit-shifting for when you need
bit-shifting semantics - for example, simulating a shift register.
>
--
Philip Potter pgp <atdoc.ic.ac. uk

So if I want to multiply an int by two I have to use a * 2 (or a + a),
but not << ? To be sure it works on every architecture (currently I
ignore the problem of the size of an int).

[Mark Bluemel]

Re: lshift &amp; rshift

iesvs@free.fr wrote:

So if I want to multiply an int by two I have to use a * 2 <snip>?

Why would you want to do it any other way?

[iesvs@free.fr]

Re: lshift &amp; rshift

On Sep 27, 10:56 am, Mark Bluemel <mark_blue...@p obox.comwrote:

ie...@free.fr wrote:

So if I want to multiply an int by two I have to use a * 2 <snip>?

>
Why would you want to do it any other way?

If << is faster, it would be great. I love to have fast software, I'm
not rich and don't buy expensive computer.

[Ian Collins]

Re: lshift &amp; rshift

iesvs@free.fr wrote:

On Sep 27, 10:56 am, Mark Bluemel <mark_blue...@p obox.comwrote:

>ie...@free.f r wrote:

>>So if I want to multiply an int by two I have to use a * 2 <snip>?

>Why would you want to do it any other way?

>
If << is faster, it would be great. I love to have fast software, I'm
not rich and don't buy expensive computer.
>

What did your profiler tell you?

--
Ian Collins.

[karthikbalaguru]

Re: lshift &amp; rshift

On Sep 27, 2:10 pm, "ie...@free .fr" <ie...@free.frw rote:

On Sep 27, 10:56 am, Mark Bluemel <mark_blue...@p obox.comwrote:
>

ie...@free.fr wrote:

So if I want to multiply an int by two I have to use a * 2 <snip>?

>

Why would you want to do it any other way?

>
If << is faster, it would be great. I love to have fast software, I'm
not rich and don't buy expensive computer.

Did you try to view the assembly of a * 2 or the assembly of a + a .
I think,a*2 and a+a must internally be doing shifting using the shift
operators if shift operators are very fast.
Am i right ?

If thatz wrong then, How do you think, a*2 or a+a will be internally ?

Karthik Balaguru

[Mark Bluemel]

Re: lshift &amp; rshift

iesvs@free.fr wrote:

On Sep 27, 10:56 am, Mark Bluemel <mark_blue...@p obox.comwrote:

>ie...@free.f r wrote:

>>So if I want to multiply an int by two I have to use a * 2 <snip>?

>Why would you want to do it any other way?

>
If << is faster, it would be great. I love to have fast software, I'm
not rich and don't buy expensive computer.

As others have already pointed out, the compiler will be more than
capable of applying this sort of optimisation if it is appropriate. In
general you should write clear, correct, expressive code and let the
compiler deal with optimisation. Any decent compiler is likely to do a
better job of optimising this sort of thing than you could.

If you have performance issues, you are generally better off looking for
better algorithms than fiddling with this sort of trivia.

If you have to get down to tuning at the code, rather than algorithm,
level then you need to profile your code as Ian has suggested. Only by
measuring will you find the bottlenecks.

[iesvs@free.fr]

Re: lshift &amp; rshift

Did you try to view the assembly of a * 2 or the assembly of a + a .

I think,a*2 and a+a must internally be doing shifting using the shift
operators if shift operators are very fast.
Am i right ?

No you're wrong. A saw that it change a * 2 by a + a. (in reality I
was looking at a * 7) and I saw something like that :
b = a + a
c = b + b
c + b + a
So after I try a * 2 and get
a + a
But it didn't use lshift.

[iesvs@free.fr]

Re: lshift &amp; rshift

What did your profiler tell you?

What is a profiler ?

[Ian Collins]

Re: lshift &amp; rshift

iesvs@free.fr wrote:

>What did your profiler tell you?

>
What is a profiler ?
>

A very useful tool that's included in any self respecting tool set.

http://en.wikipedia.org/wiki/Performance_analysis

--
Ian Collins.

[iesvs@free.fr]

Re: lshift &amp; rshift

On Sep 27, 11:28 am, Ian Collins <ian-n...@hotmail.co mwrote:

ie...@free.fr wrote:

What did your profiler tell you?

>

What is a profiler ?

>
A very useful tool that's included in any self respecting tool set.
>

http://en.wikipedia.org/wiki/Performance_analysis

>
--
Ian Collins.

I never use these kind of stuffs.

[Ian Collins]

Re: lshift &amp; rshift

iesvs@free.fr wrote:

On Sep 27, 11:28 am, Ian Collins <ian-n...@hotmail.co mwrote:

>ie...@free.f r wrote:

>>>What did your profiler tell you?
>>What is a profiler ?

>A very useful tool that's included in any self respecting tool set.
>>
>http://en.wikipedia.org/wiki/Performance_analysis
>>

>
I never use these kind of stuffs.
>

Then you shouldn't be fiddling with micro-optimisations.

--
Ian Collins.

[pete]

Re: lshift &amp; rshift

Richard Heathfield wrote:

>
iesvs@free.fr said:
>

Hello guys,
every time a rode a doc or a book about the language C I saw that
operators << and >exist. But each time they said that << translate
the digit

>
Each bit is moved one place to the left,
the leftmost bit falls off the end
into the bit bucket (which should be emptied regularly),
and the rightmost bit is set to 0.
>

to the left (and >...)

>
Beware signed types when shifting. The danger is particularly - er -
dangerous when right-shifting negative numbers.
What happens to the sign bit?
(Answer: it depends on the implementation. )
>

but no one said if a << 1 mean
every time a * 2,

>
Not every time, no.
For example, consider a system where CHAR_BIT is 8.
>
unsigned char ch = 1;
>
ch <<= 1; /* ch is now 2 */
ch <<= 1; /* ch is now 4 */
ch <<= 1; /* ch is now 8 */
ch <<= 1; /* ch is now 16 */
ch <<= 1; /* ch is now 32 */
ch <<= 1; /* ch is now 64 */
ch <<= 1; /* ch is now 128 */
ch <<= 1; /* ch is now 0, even though 2 * 128 is 256. */

That example doesn't show a difference
between shifting and a doubling.

The result of each shift
is the same as what you would get from multiplication.

CHAR_BIT is 8

unsigned char ch = 1;

ch *= 2; /* ch is now 2 */
ch *= 2; /* ch is now 4 */
ch *= 2; /* ch is now 8 */
ch *= 2; /* ch is now 16 */
ch *= 2; /* ch is now 32 */
ch *= 2; /* ch is now 64 */
ch *= 2; /* ch is now 128 */
ch *= 2; /* ch is now 0, even though 2 * 128 is 256. */

--
pete