Left Shift / Right Shift Operators

[Santosh Nayak]

Hi,
Is there any way to catch the losing bit occurring due to Right Shift
Operator ?

e.g
int a = 5 ;
a = a >1 ; // // a is now 2 and the least significant bit is lost //
//

I want this solution for the question:
"How to find if the number is even or odd using only "<<" or/and ">>"
operators ?"

[Harald van Dijk]

Richard Heathfield wrote:

pete said:

Santosh Nayak wrote:

>
Hi,
Is there any way to catch the losing bit occurring due to Right Shift
Operator ?

e.g
int a = 5 ;
a = a >1 ; // // a is now 2 and the least significant bit is lost //
//

I want this solution for the question:
"How to find if the number is even or odd using only "<<" or/and ">>"
operators ?"

b = 0 a ? -a : a;
printf("%d is %s.\n", a, b - (b >1 << 1) ? "odd" : "even");

Your use of these operators:

= ?: -

violates the OP's constraint.

Here is a more accurate (although admittedly not terribly informative) way
to find if the number is even or odd:

int a = 5;
puts("Yes.");
/* oops, I forgot to use the << and >operators, so I'll do that now */
a >1;
a << 1;

Your use of the function call operator violates the OP's constraint.

Here's a more accurate /and/ informative, but not entirely portable,
way to find if the number is even or odd:

int main(int argc, char *argv[]) {
struct { unsigned b : 1; } s = { argc };
return s.b;
}

Pointless use of << and >can be added here too.

[pete]

Richard Heathfield wrote:

>
pete said:

Santosh Nayak wrote:

>
Hi,
Is there any way to catch the losing bit
occurring due to Right Shift Operator ?

e.g
int a = 5 ;
a = a >1 ;

I want this solution for the question:
"How to find if the number is even or odd using
only "<<" or/and ">>"
operators ?"

b = 0 a ? -a : a;
printf("%d is %s.\n", a, b - (b >1 << 1) ? "odd" : "even");

Your use of these operators:

= ?: -

violates the OP's constraint.

OP restated his constraint in another strand of this thread,
as only applying to bitwise operators:

"Perhaps, i was not clear about my question.
I meant we are not supposed to use any other bitwise operators
other than "<<" or ">>"."

http://groups-beta.google.com/group/...a68f5071ead837

--
pete

[Random832]

2006-11-30 <11************ **********@14g2 000cws.googlegr oups.com>,
Santosh Nayak wrote:

On Nov 30, 11:54 am, "Arthur J. O'Dwyer" <ajonos...@andr ew.cmu.edu>
wrote:

> int a = 5;
int lost_bit = a & 1;

Perhaps, i was not clear about my question.
I meant we are not supposed to use any other bitwise operators other
than "<<" or ">>".
"&" operator is not allowed.

You don't need to catch the shifted-off bit to detect if a number is odd
or even, but in order to avoid doing so you need to use _both_ << and >>

[Richard Heathfield]

Harald van D?k said:

Richard Heathfield wrote:

>pete said:

Santosh Nayak wrote:

Hi,
Is there any way to catch the losing bit occurring due to Right Shift
Operator ?

e.g
int a = 5 ;
a = a >1 ; // // a is now 2 and the least significant bit is lost //
//

I want this solution for the question:
"How to find if the number is even or odd using only "<<" or/and ">>"
operators ?"

b = 0 a ? -a : a;
printf("%d is %s.\n", a, b - (b >1 << 1) ? "odd" : "even");

Your use of these operators:

= ?: -

violates the OP's constraint.

Here is a more accurate (although admittedly not terribly informative)
way to find if the number is even or odd:

int a = 5;
puts("Yes.") ;
/* oops, I forgot to use the << and >operators, so I'll do that now */
a >1;
a << 1;

Your use of the function call operator violates the OP's constraint.

<grinI wondered if anyone would pick that up. I don't think the OP would
be too concerned about an output call, though. Okay, perhaps not /that/
output!

Here's a more accurate /and/ informative, but not entirely portable,
way to find if the number is even or odd:

int main(int argc, char *argv[]) {
struct { unsigned b : 1; } s = { argc };
return s.b;
}

Use of the structure member operator violates the OP's constraint. :-)

Pointless use of << and >can be added here too.

Thanks. << << << ><< >>

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

[Richard Heathfield]

pete said:

Richard Heathfield wrote:

>>
pete said:

<snip>

>
b = 0 a ? -a : a;
printf("%d is %s.\n", a, b - (b >1 << 1) ? "odd" : "even");

Your use of these operators:

= ?: -

violates the OP's constraint.

OP restated his constraint in another strand of this thread,
as only applying to bitwise operators:

"Perhaps, i was not clear about my question.
I meant we are not supposed to use any other bitwise operators
other than "<<" or ">>"."

http://groups-beta.google.com/group/...a68f5071ead837

In which case it's trivial.

int is_odd(unsigned long foo)
{
int i = (2 << 1) >1;

return foo % i;
}

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

[Harald van Dijk]

Richard Heathfield wrote:

Harald van D?k said:

Richard Heathfield wrote:

pete said:

Santosh Nayak wrote:

Hi,
Is there any way to catch the losing bit occurring due to Right Shift
Operator ?

e.g
int a = 5 ;
a = a >1 ; // // a is now 2 and the least significant bit is lost //
//

I want this solution for the question:
"How to find if the number is even or odd using only "<<" or/and ">>"
operators ?"

b = 0 a ? -a : a;
printf("%d is %s.\n", a, b - (b >1 << 1) ? "odd" : "even");

Your use of these operators:

= ?: -

violates the OP's constraint.

Here is a more accurate (although admittedly not terribly informative)
way to find if the number is even or odd:

int a = 5;
puts("Yes.");
/* oops, I forgot to use the << and >operators, so I'll do that now */
a >1;
a << 1;

Your use of the function call operator violates the OP's constraint.

<grinI wondered if anyone would pick that up. I don't think the OP would
be too concerned about an output call, though. Okay, perhaps not /that/
output!

Here's a more accurate /and/ informative, but not entirely portable,
way to find if the number is even or odd:

int main(int argc, char *argv[]) {
struct { unsigned b : 1; } s = { argc };
return s.b;
}

Use of the structure member operator violates the OP's constraint. :-)

Good point. :-)

#include <tgmath.h>
int main(int argc, char *argv[]) {
_Bool b = fmod(argc, 2); /* macro invocation, not a function call
*/
return b;
}

(The _Bool variable is because otherwise, rounding errors cause
problems on my system.)

Pointless use of << and >can be added here too.

Thanks. << << << ><< >>

[CBFalconer]

Richard Heathfield wrote:

pete said:

>Richard Heathfield wrote:

>>pete said:

<snip>

>>>>
b = 0 a ? -a : a;
printf("%d is %s.\n", a, b - (b >1 << 1) ? "odd" : "even");

Your use of these operators:

= ?: -

violates the OP's constraint.

OP restated his constraint in another strand of this thread,
as only applying to bitwise operators:

"Perhaps, i was not clear about my question.
I meant we are not supposed to use any other bitwise operators
other than "<<" or ">>"."

http://groups-beta.google.com/group/...a68f5071ead837

In which case it's trivial.

int is_odd(unsigned long foo)
{
int i = (2 << 1) >1;

return foo % i;
}

Now that you have started doing his homework for him, I suggest:

int is_odd(unsigned long foo) {

return foo - ((foo >1) << 1);
}

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net>