From: Josef 'Jupp' Schugt <jupp@...>
Date: 2003-06-01T04:04:51+09:00
Subject: Re: expandtabs

Saluton!


* Steven Shaw, 2003-05-30, 21:05 UTC:
> The methods for expanding tabs in the Ruby FAQ don't seem to work.

> def method1(a)
>   1 while a.sub!(/(^[^\t]*)\t(\t*)/){$1+' '*(8-$1.size%8+8*$2.size)}
>   a
> end
> 
> def method2(a)
>   1 while a.sub!(/\t(\t*)/){' '*(8-$~.begin(0)%8+8*$1.size)}
>   a
> end
> 
> def method3(a)
>   a.gsub!(/([^\t]{8})|([^\t]*)\t/n){[$+].pack("A8")}
>   a
> end
> 
> def println(s)
>   print s, "\n"
> end
> 
> string = ""
> (0..10).each {|n|
>   string << "blah\t" + (" " * n) + "blah\n"
> }
> 
> println(method1(string.clone))
> println(method2(string.clone))
> println(method3(string.clone))

I did already answer that in a PM. The above works correct but does
not have the intended result. The problem is that the sample string
does contain "\n" characters.

The easiest solution would be using

string.split("\n").each{|x| println(method1(x))}
string.split("\n").each{|x| println(method2(x))}
string.split("\n").each{|x| println(method3(x))}

in place of the  given calls of println. Nevertheless it would be
better to reset the length counting for each "\n". The following is a
quick'n'dirty hack that does this in a way that is far from optimal.

def method4(a)
  b = ''
  l = 1
  0.upto(a.length) { |i|
    case a[i]
      when 9
	b += ' '
	l += 1
	while l % 8 != 0
	  b += ' '
	  l += 1
	end
      when 10
	b += a[i..i]
	l = 1
      else
	b += a[i..i]
	l += 1
    end
  }
  return b
end

A shorter solution is:

def method1(a)
  a.split("\n").each {|b|
    1 while b.sub!(/(^[^\t]*)\t(\t*)/){$1+' '*(8-$1.size%8+8*$2.size)}
    b
  }.join("\n")
end

The equivalents for method2 and method3 are obvious. Does anybody
have a better solution that does not require splitting and re-joining
(which is time-consuming)?

Gis,

Josef 'Jupp' Schugt
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