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regexlex

Lex: match ignore space

I have a work to recognize hex number, my problem is how to ignore space, but not allow any character before.
like this: 

0x7f6e ---->match,and print"0x7f6e"
    0X2146 ---->match,and print"0X21467"
acns0x8972 ----> not match

my work now:

hex     \s*0[X|x][0-9a-fA-f]{1,4}(^.)*(\n)

{hex}   { ECHO;}
.|\n    {}

and it print: 

0x7f6e
    0X2146

how can i print it without space? like this:

0x7f6e
0X2146
Solution

  • I got a working version which should do what you expect:

    %{
#include <ctype.h>
#include <stdio.h>
%}

%%

^[ \t]*0[Xx][0-9a-fA-f]{1,4}(.*)$ {
  /* skip spaces at begin of line */
  const char *bol = yytext;
  while (isspace((unsigned char)*bol)) ++bol;
  /* echo rest of line */
  puts(bol);
}

.|\n { }

%%

int main(int argc, char **argv) { return yylex(); }

int yywrap() { return 1; }

    Notes:

    1. \s seems to be unsupported (at least in my version 2.6.3 of flex). I replaced it by [ \t]. Btw. \s usually matches also carriage return, newline, formfeed what's not intended in my case.

    2. (^.)* replaced by (.*). (I didn't understand the intention of the original one. Mistake?)

    3. I added a ^ at begin of 1st pattern so that pattern is attached to begin of line.

    4. I replaced \n at the end of hex line with $. The puts() function adds a newline to output. (Newlines are always matched by 2nd rule and thus skipped.)

    5. I replaced ECHO; with some C code to (1st) remove spaces at begin of line and (2nd) output the rest of line to standard output channel.

    Compiled and tested in cygwin on Windows 10 (64 bit):

    $ flex --version
flex 2.6.3

$ flex -o test-hex.c test-hex.l ; gcc -o test-hex test-hex.c

$ echo "
0x7f6e                                              
    0X2146
acns0x8972
" | ./test-hex
0x7f6e
0X2146

$

    Note: I used echo to feed your sample data via pipe into standard input channel of test-hex.