Regex scan/tokenize from right to left order?

Ruby
1

I’m trying to evaluate an arithmetic operation of prefix type,
“add(multiply(123,432),subtract(add(19,21),124))”, can you make it so
the regex engine scans it backwards?

I’m trying to isolate each word and manipulate it without using split()
and creating another array of o(n) space.

I can parse the string from end to start by character and tokenize it
using string index, but there must be someway to do this using regex
efficiently.

I know that split() creates an array that can be traversed backwards,
but is there anyway to bypass the extra space required by split()? (Does
the ruby interpreter/compiler somehow handle it?)

Ultimately, is there anyway to make scan() parse a string backwards?

Thanks

2

You need to show what code you have currently.

3

On Sun, Feb 9, 2014 at 12:43 PM, Tonton S. [email protected]
wrote:

I’m trying to evaluate an arithmetic operation of prefix type,
“add(multiply(123,432),subtract(add(19,21),124))”, can you make it so
the regex engine scans it backwards?

Note Ruby’s regexp engine since 1.9 can parse nested structures (i.e.
non regular languages but context free languages). You’ll find some
hints here
http://www.geocities.jp/kosako3/oniguruma/doc/RE.txt

Ultimately, is there anyway to make scan() parse a string backwards?

No. You can reverse the String but then you also have to reverse the
regexp which is not trivial for all cases.

Kind regards

robert

4

Tonton S. wrote in post #1136122:

I’m trying to evaluate an arithmetic operation of prefix type,
“add(multiply(123,432),subtract(add(19,21),124))”, can you make it so
the regex engine scans it backwards?

why scans backward ?

s=“add(multiply(2,4),substract(add(add(add(1,2),9),add(2,1)),1))”

l=s.split(/([)(,-])/).reject {|a| a==’(’ || a==’)’ || a==’,’ || a==""}
p l
reg=[]

def add(a,b) (a.to_i+b.to_i).to_s end
def multiply(a,b) (a.to_i*b.to_i).to_s end
def substract(a,b) (b.to_i-a.to_i).to_s end

def is_num(a) a=~/^-?\d+$/ end
def is_op(a) a=~/^[a-z]+$/ end

loop do
if is_num(l[-1]) && is_num(l[-2]) && is_op(l[-3])
a,b,op=l.pop,l.pop,l.pop
l.push send(op,a,b)
end
if is_num(l[-1]) && is_num(l[-2]) && is_num(l[-3])
reg << l.pop
end
if is_num(l[-1]) && is_op(l[-2]) && reg.size>0
l.push reg.pop
end
puts “#{l.inspect} || #{reg.inspect}”
break if l.size==1 && reg.size==0
end
p l.last

Perhaps you mix lexical analysis with syntax analysis and evaluator ?