Just because you *can* create a malformed string with no delimiter does not mean that my statement about proper C behavior is untrue.
It is a true statement that "the z architecture stores integers in big-endian form." Nonetheless, I *can* create a little-endian integer on z. That does not make the statement untrue. BTW, C++ gives an error for the s6[6] = "wombat" case. Charles -----Original Message----- From: IBM Mainframe Discussion List [mailto:[email protected]] On Behalf Of Paul Gilmartin Sent: Tuesday, September 04, 2012 6:47 AM To: [email protected] Subject: Re: The IBM zEnterprise EC12 announcment On Tue, 4 Sep 2012 06:22:43 -0700, Charles Mills wrote: > >char[] delimits strings with '\0' in every implementation in the world, I >think. > It depends. With gcc: char a6[ 6 ] = { 'w', 'o', 'm', 'b', 'a', 't' }; char s6[ 6 ] = "wombat"; both compile with no error message, whereas: char s5[ 5 ] = "wombat"; gives: wombat.c:6: warning: initializer-string for array of chars is too long so, apparently the \0 terminator is not mandatory. - ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to [email protected] with the message: INFO IBM-MAIN
