bitwise shift help

K

Kvetch Kvetch

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
 
K

Kvetch Kvetch

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

Thanks
 
A

Axel

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
 
C

Caleb Clausen

Kvetch said:
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]
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?

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?
 
K

Kvetch Kvetch

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 said:
Kvetch said:
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?
 
B

Bertram Scharpf

Hi,

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

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 of `String#notempty?',
`Array#notempty?', and `Hash#notempty?'.

`Bertram::NotEmpty#insists?' still returns `true'.

Bertram
 

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