Bitwise shift help

Ruby
1

Hello, I am new to Ruby and was wondering if there is an easier/faster
way to determine if a bit is set in a number. For example,
If my number is 1011000010 and I want to check to see if 3rd bit is set
so
1011000010
1000010000

  0000 - So 3rd bit is not set

My code is

y = 1 << 3
x = 0b1011000010

if ( x & y ) == 1
puts “The 3rd bit is set”
else
puts “The 3rd bit is not set”
end

Is there a better/faster method of achieving this bit check?
Thank you

2

Actually I suppose the bit check should read
if ( x & y ) >= 1
Or I am still totally off.

Thanks

Kvetch K. wrote:

Hello, I am new to Ruby and was wondering if there is an easier/faster
way to determine if a bit is set in a number. For example,
If my number is 1011000010 and I want to check to see if 3rd bit is set
so
1011000010
1000010000

  0000 - So 3rd bit is not set

My code is

y = 1 << 3
x = 0b1011000010

if ( x & y ) == 1
puts “The 3rd bit is set”
else
puts “The 3rd bit is not set”
end

Is there a better/faster method of achieving this bit check?
Thank you

3

Maybe this:

Fixnum[n]
Bit Reference—Returns the nth bit in the binary representation of
fix, where fix[0] is the least significant bit.

–Axel

4

On 1/3/10, Kvetch K. [email protected] wrote:

Kvetch K. wrote:

Hello, I am new to Ruby and was wondering if there is an easier/faster
way to determine if a bit is set in a number. For example,
[snip]

Actually I suppose the bit check should read
if ( x & y ) >= 1
Or I am still totally off.

There’s nonzero?, which ought to be a tad faster than >=1. So, write
this instead:

if ( x & y ).nonzero?

Or, if you incorporate Axel’s suggestion to use Fixnum#[] as well, then:

if x[3].nonzero?

5

Hi,

Am Montag, 04. Jan 2010, 13:38:08 +0900 schrieb Kvetch K.:

y = 1 << 3
x = 0b1011000010

if ( x & y ) == 1
puts “The 3rd bit is set”
else
puts “The 3rd bit is not set”
end

Is there a better/faster method of achieving this bit check?
Thank you

As in SQL, 0 does not yield false. SQL’s NULL is different from
0 and from ‘’. So is in Ruby `nil’ different from 0, from “”, and
from [], and further from {}. In C you’re free to write

if (num % 7) { … }

That won’t work in Ruby.

Numeric#nonzero?' has been mentioned. Of course, I cannot resist and just mention my beloved proposal ofString#notempty?’,
Array#notempty?', andHash#notempty?’.

Bertram::NotEmpty#insists?' still returnstrue’.

Bertram

6

Awesome thank you Caleb and Axel.
I had tried the fixnum but it wasn’t working and then it occurred to me
that it was being read in as a string. Thank you both for the quick
answers. I appreciate the help.

Caleb C. wrote:

On 1/3/10, Kvetch K. [email protected] wrote:

Kvetch K. wrote:

Hello, I am new to Ruby and was wondering if there is an easier/faster
way to determine if a bit is set in a number. For example,
[snip]

Actually I suppose the bit check should read
if ( x & y ) >= 1
Or I am still totally off.

There’s nonzero?, which ought to be a tad faster than >=1. So, write
this instead:

if ( x & y ).nonzero?

Or, if you incorporate Axel’s suggestion to use Fixnum#[] as well, then:

if x[3].nonzero?