I know this . I get x'5f' with ibm-037, but x'6b' otherwise. The problem is in 
loading/conuertng   za

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-----Original message-----
From: Charles Mills <charl...@mcn.org>
To: IBM-MAIN@LISTSERV.UA.EDU
Sent: Fri, Dec 7, 2012 20:20:22 GMT+00:00
Subject: Re: How to load logical load x'ac'

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:IBM-MAIN@LISTSERV.UA.EDU] On
Behalf Of Ze'ev Atlas
Sent: Friday, December 07, 2012 11:08 AM
To: IBM-MAIN@LISTSERV.UA.EDU
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. 

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