Convert a n binary number to /n

[Zangief Ief]

Hi,

If my number is 1111, I would like to had 0000.
If my number is 1001, I would like to had 0110.
If my number is 0011, I would like to had 1100.

Of cours I know we can use XOR operator, but I would like to allow Ruby
to do it more simply, just by using a function as simple as possible. I
a sure this kind of function is yet setup in Ruby, but I don't know what
is his name.

Thank you for help.

Zang

 

[Xavier Noria]

If my number is 1111, I would like to had 0000.
If my number is 1001, I would like to had 0110.
If my number is 0011, I would like to had 1100.

Of cours I know we can use XOR operator, but I would like to allow
Ruby
to do it more simply, just by using a function as simple as
possible. I
a sure this kind of function is yet setup in Ruby, but I don't know
what
is his name.

A solution:

irb(main):003:0> "0011".tr("01", "10")
=> "1100"

-- fxn

 

[Suraj Kurapati]

irb(main):003:0> "0011".tr("01", "10")
=> "1100"

Or, if you're working with integers (as opposed to strings), then you
can use the bitwise negation operator (~) and avoid the overhead of
string conversions altogether:
=> 10

 

[Zangief Ief]

Thanks

 

[Todd Burch]

If my number is 1111, I would like to had 0000.
If my number is 1001, I would like to had 0110.
If my number is 0011, I would like to had 1100.
Thanks

sprintf("%04d", 1111- 1111) -> 0000
sprintf("%04d", 1111- 1011) -> 0100
sprintf("%04d", 1111- 1001) -> 0110
sprintf("%04d", 1111- 0011) -> 1102 (oops - it treats the second
value as octal)
sprintf("%04d", 1111- "0011".to_i) -> 1100

Todd