What is the C language operator you are trying to use? Bitwise XOR? The "usual" C meaning of the circumflex?
On my PC I key it in and it appears as a circumflex. Now, looking at the code on zOS with the ISPF editor (which I almost never do) I see code point 5F which my emulator is rendering as a logical not. If I configure my emulator for code page 1047 then the emulator renders code point 5F as a circumflex. Would that solve your problem? You have to understand there are a lot of layers here. It is not correct to say that your ASCII circumflex has *become* a logical not. It has become code point 5f, which the C compiler apparently understands as bitwise logical not. A particular emulator or printer might render the EBCDIC code point in any of a variety of ways. Charles -----Original Message----- From: IBM Mainframe Discussion List [mailto:[email protected]] On Behalf Of Ze'ev Atlas Sent: Friday, December 07, 2012 11:08 AM To: [email protected] Subject: Re: How to load logical load x'ac' The z/OS native C compiler does not compile the source code when I do circumflex, only logical not. I may resort to the digraph but that would make the code less readable. However, I would appreciate if you show here the JCL that compiles source code with circumflex on z/OS, maybe with a short 3 line program with that circumflex. My assumption is that you have some compile option that I do not know about (and could not find in the manual) that allow C to understand the circumflex. ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to [email protected] with the message: INFO IBM-MAIN
