How do I increment/decrement a character in Ruby for all possible values?

Question

I have a string that is one character long and can be any possible character value:

irb(main):001:0> \"\\x0\"
=> \"\\u0000\"

I though

Answer 1

You could use the String#next method.

Answer 2

Depending on what the possible values are, you can use String#next:

"\x0".next
# => "\u0001"

Or, to update an existing value:

c = "\x0"
c.next!

This may well not be what you want:

"z".next
# => "aa"

The simplest way I can think of to increment a character's underlying codepoint is this:

c = 'z'
c = c.ord.next.chr
# => "{"

Decrementing is slightly more complicated:

c = (c.ord - 1).chr
# => "z"

In both cases there's the assumption that you won't step outside of 0..255; you may need to add checks for that.

Answer 3

I think the most elegant method (for alphanumeric chars) would be:

"a".tr('0-9a-z','1-9a-z0')

which would loop the a through to z and through the numbers and back to a.

I reread the question and see, that my answer has nothing to do with the question. I have no answer for manipulationg 8-bit values directly.

Answer 4

You cannot do:

"\x0" += 1

Because, in Ruby, that is short for:

"\x0" = "\x0" + 1

and it is a syntax error to assign a value to a string literal.

However, given an integer n, you can convert it to a character by using pack. For example,

[97].pack 'U' # => "a"

Similarly, you can convert a character into an integer by using ord. For example:

[300].pack('U').ord # => 300

With these methods, you can easily write your own increment function, as follows:

def step(c, delta=1)
  [c.ord + delta].pack 'U'
end

def increment(c)
  step c, 1
end

def decrement(c)
  step c, -1
end

If you just want to manipulate bytes, you can use String#bytes, which will give you an array of integers to play with. You can use Array#pack to convert those bytes back to a String. (Refer to documentation for encoding options.)