Shell Programming and Scripting

I am having file xyz.log
Its content is like this

Code:

int main()
{
  d;

  #ifdef
      e;
      f;
  #else 
      g;


  #ifdef
      h.
   #else 
      i;


  RET_OK;
}

-------------------------------
I need to delete the lines starting from
#ifdef to #else

when i am trying this command

Code:

s/#ifdef(.|n)*#else//;

It is deleting like this
--------------

Code:

int main ();
{
d;
i;
RET_OK;
}

-------------

It is deleting "g" also

It is searching from #ifdef until the last #else , and replacing.
I need to delete from #ifdef to the nearest match "#else"
How to do that

what will be the best script command in perl to do this

Moderator's Comments:

How to use code tags

The range expression of awk

The pattern1, pattern2 form of an expression is called a range pattern. It matches all input records starting with a record that matches pattern1, and continuing until a record that matches pattern2, inclusive.

may be a better choice. If you have:

Code:

awk '/^#ifdef/,/^#else/ { print; }' filename

you will get:

Code:

#ifdef
e;
f;
#else
#ifdef
h.
#else

which is the lines you want to remove. You cannot negate the pattern:

Code:

awk '! /^#ifdef/,/^#else/ { print; }' filename

instead, you have to do something like:

Code:

awk '{ f = 1; } /^#ifdef/,/^#else/ { f = 0; } f { print; }' filename

which will print the lines not inside the pattern.
Hope this helps.

The Perl way.

Code:

perl -i.bak -pe '(/^#ifdef/../^#else/)&&s/$_//g' inputfile

Also, this one-liner will create a backup of the original input file with '.bak' in its name.