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The memory allocated by the 'grub_parser_split_cmdline()' is leaked.

From: Hitoshi Ozeki
Subject: The memory allocated by the 'grub_parser_split_cmdline()' is leaked.
Date: Tue, 10 Apr 2007 21:46:17 +0900

Hello, all.

Please see line 204 and 209 on the 'grub-1.95/kern/parser.c'.

This code allocates memory. it stores arguments.

  /* Reserve memory for the return values.  */
  args = grub_malloc (bp - buffer);
  if (! args)
    return grub_errno;
  grub_memcpy (args, buffer, bp - buffer);
  *argv = grub_malloc (sizeof (char *) * (*argc + 1));
  if (! *argv)
      grub_free (args);
      return grub_errno;

And this code is called from 'grub_normal_do_completion ()' and 
'grub_enter_rescue_mode ()'. But I cannot find the codes release
its memory. Maybe leaks it.

 Hitoshi Ozeki address@hidden

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