[Bug c++/16396] New: cout of volatile char * prints funny

["ramsdell at mitre dot org" <gcc-bugzilla at gcc dot gnu dot org>]

Are volatile C strings printed in C++ by special rules?  I get 1 no matter
what is in the string.  See the comments at the end of the following C++ program.
                                                                                
$ cat vcp.cc
#include <cstdio>
#include <iostream>
 
int main(int argc, char **argv)
{
  volatile char str[] = { 'a', 'b', '\0' };
  std::printf("%s\n", str);
  std::cout << str << "\n";
  return 0;
}
 
/*
 * $ uname -a
 * Linux localhost.localdomain 2.6.5-1.358 #1 Sat May 8 09:04:50 EDT 2004 i686
i686 i386 GNU/Linux
 * $ gcc --version
 * gcc (GCC) 3.3.3 20040412 (Red Hat Linux 3.3.3-7)
 * Copyright (C) 2003 Free Software Foundation, Inc.
 * This is free software; see the source for copying conditions.  There is NO
 * warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
 *
 * $ make vcp
 * g++     vcp.cc   -o vcp
 * $ ./vcp
 * ab
 * 1
 * $
 */
$

-- 
           Summary: cout of volatile char * prints funny
           Product: gcc
           Version: 3.3.3
            Status: UNCONFIRMED
          Severity: normal
          Priority: P2
         Component: c++
        AssignedTo: unassigned at gcc dot gnu dot org
        ReportedBy: ramsdell at mitre dot org
                CC: gcc-bugs at gcc dot gnu dot org
 GCC build triplet: i386-redhat-linux
  GCC host triplet: i386-redhat-linux
GCC target triplet: i386-redhat-linux


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=16396