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Top Forums Shell Programming and Scripting Help building a variable string from a keyword - character replacements! Post 303031039 by Chubler_XL on Wednesday 20th of February 2019 08:33:19 PM
Old 02-20-2019
Another approach using substr:

Code:
awk '
{ 
  printf "KEYWORD=\""
  for (i=0; i <= length($0) - 2; i++)
    for(j=i + 1; j <= length($0) - 1; j++)
      printf "%s%s", \
          substr($0, 0, i) "_"  \
          substr($0, i + 2, j - i - 1) "_"  \
          substr($0, j + 2),  \
          i == (length($0) - 2) ? "\"" : " "
   print ""
}'


Last edited by Chubler_XL; 02-20-2019 at 09:50 PM..
 

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Perl::Critic::Policy::BuiltinFunctions::ProhibitLvalueSuUser(Contributed Perl DocumPerl::Critic::Policy::BuiltinFunctions::ProhibitLvalueSubstr(3)

NAME
Perl::Critic::Policy::BuiltinFunctions::ProhibitLvalueSubstr - Use 4-argument "substr" instead of writing "substr($foo, 2, 6) = $bar". AFFILIATION
This Policy is part of the core Perl::Critic distribution. DESCRIPTION
Conway discourages the use of "substr()" as an lvalue, instead recommending that the 4-argument version of "substr()" be used instead. substr($something, 1, 2) = $newvalue; # not ok substr($something, 1, 2, $newvalue); # ok The four-argument form of "substr()" was introduced in Perl 5.005. This policy does not report violations on code which explicitly specifies an earlier version of Perl (e.g. "use 5.004;"). CONFIGURATION
This Policy is not configurable except for the standard options. SEE ALSO
"substr" in perlfunc (or "perldoc -f substr"). "4th argument to substr" in perl5005delta AUTHOR
Graham TerMarsch <graham@howlingfrog.com> COPYRIGHT
Copyright (c) 2005-2011 Graham TerMarsch. All rights reserved. This program is free software; you can redistribute it and/or modify it under the same terms as Perl itself. perl v5.16.3 2014-06-09 Perl::Critic::Policy::BuiltinFunctions::ProhibitLvalueSubstr(3)
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