Why is regexp "\Amax_repeats=(\d+)" not equal to "^max_repeats=([0-9]+)" ?

[Martin Elzen]

Hi everyone.

I'm trying to learn how to program in Ruby, and yesterday I came across a
Regexp issue I can't figure out. I want to parse a commandline argument of
the form: max_repeats=<integer>

After studying the 'pick-axe' book, I came up with the line:
regexp = Regexp.new("\Amax_repeats=(\d+)")

but the above regexp fails to match with correct input. When I changed that
line to the following:
regexp = Regexp.new("^max_repeats=([0-9]+)")

it *does* match correctly with the proper input.

What I would like to know is, why doesn't the first Regexp work? (FWIW, I'm
using Ruby 1.8.1 on Win2003 server.)

Sincerely,
Martin

PS a complete program with my regexp matching code is as follows:

#define globals

$max_repeat_count_default = 5




def get_max_repeats_value
result = $max_repeat_count_default

if not ARGV.empty?

#regexp = Regexp.new("\Amax_repeats=(\d+)")
#@@@
regexp = Regexp.new("^max_repeats=([0-9]+)")

matchdata = candidate_val = match_val = nil

ARGV.each do |arg|
matchdata = regexp.match(arg)

if matchdata != nil
match_val = matchdata[1].to_i

if candidate_val == nil
candidate_val = match_val
else
candidate_val = match_val if match_val > candidate_val
end

end

end

result = candidate_val if candidate_val != nil and candidate_val > 0

end

result
end



begin
max_repeats = get_max_repeats_value
puts "max_repeats=" + max_repeats.to_s

rescue Exception => ex
print "Exception exception: " + ex.message +
",\nbacktrace:\n" + ex.backtrace.join("\n")

end

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[Simon Strandgaard]

I'm trying to learn how to program in Ruby, and yesterday I came across a
Regexp issue I can't figure out. I want to parse a commandline argument of
the form: max_repeats=<integer>

After studying the 'pick-axe' book, I came up with the line:
regexp = Regexp.new("\Amax_repeats=(\d+)")

but the above regexp fails to match with correct input. When I changed that
line to the following:
regexp = Regexp.new("^max_repeats=([0-9]+)")

it *does* match correctly with the proper input.

Watch out about escaping inside double quoted strings..
observe the difference

irb(main):001:0> re = Regexp.new("\Ax")
=> /Ax/
irb(main):002:0> re.match("xxx").to_a
=> []
irb(main):003:0> re = Regexp.new('\Ax')
=> /\Ax/
irb(main):004:0> re.match("xxx").to_a
=> ["x"]
irb(main):005:0>


Does this help you ?

 

[ts]

M> regexp = Regexp.new("\Amax_repeats=(\d+)")
^^ ^^

You want \\ rather than \

svg% ruby -e 'p Regexp.new("\Amax_repeats=(\d+)")'
/Amax_repeats=(d+)/
svg%

svg% ruby -e 'p Regexp.new("\\Amax_repeats=(\\d+)")'
/\Amax_repeats=(\d+)/
svg%

ou use simple quote, i.e. ''


Guy Decoux

 

[Robert Klemme]

Hi everyone.

I'm trying to learn how to program in Ruby, and yesterday I came across a
Regexp issue I can't figure out. I want to parse a commandline argument of
the form: max_repeats=<integer>

After studying the 'pick-axe' book, I came up with the line:
regexp = Regexp.new("\Amax_repeats=(\d+)")

You're running into a quoting issue here. See the subtle differences:

irb(main):003:0> Regexp.new("\Amax_repeats=(\d+)")
=> /Amax_repeats=(d+)/
irb(main):004:0> Regexp.new("\\Amax_repeats=(\\d+)")
=> /\Amax_repeats=(\d+)/
irb(main):005:0> %r(\Amax_repeats=(\d+))
=> /\Amax_repeats=(\d+)/
irb(main):006:0> /\Amax_repeats=(\d+)/
=> /\Amax_repeats=(\d+)/
irb(main):007:0> Regexp.new('\Amax_repeats=(\d+)')
=> /\Amax_repeats=(\d+)/

Your code (the first line) looses the backslash ("\") because it is the
escape char in the double quoted string. You can use single quoted string
(last line) but in this case I'd prefer any of the other methods since the
string is constant anyway.

but the above regexp fails to match with correct input. When I changed that
line to the following:
regexp = Regexp.new("^max_repeats=([0-9]+)")

it *does* match correctly with the proper input.

What I would like to know is, why doesn't the first Regexp work? (FWIW, I'm
using Ruby 1.8.1 on Win2003 server.)

As said, the regexps are not near to identical. Your expression really
starts with an uppercase "a" where you wanted it to start with "\A" which
is the "start of string" anchor.

Kind regards

robert