I tried the simplest solution of casting ASXB to a CHAR * since I am not
familiar with templates and basically got the same value as before.
I will try some of the other solutions .
Thanks everyone
$ jn2
ASCB = FB7A00
ASXBC = FB7BB0
********************
int *PSA;
int *ASCB;
int *ASXB=nullptr;
char *ASXBC;
char jobnameÝ9¨;
PSA=(int *)0x224; /* address of PSAAOLD */
ASCB=(int *)*PSA;
printf("ASCB = %X\n",ASCB);
ASXBC= (char *)(ASCB + 0x6c);
printf("ASXBC = %X\n",ASXBC);
-----Original Message-----
From: IBM Mainframe Discussion List [mailto:[email protected]] On Behalf
Of Paul Gilmartin
Sent: Thursday, February 08, 2018 12:39 PM
To: [email protected]
Subject: Re: Silly C problem adding hex 6C
On Thu, 8 Feb 2018 10:22:40 -0600, Allan Kielstra wrote:
>The size of a char in 1 byte. Try
>
>(char *) ASXB = (char *) ASCB + 0x6c; /* lazy version */
>
ANSI says a cast may not be used as an L-value. IBM's C compiler
enforces this. I once did something like (IIRC?):
*(char * *) ASXB += 0x6c;
My head hurts.
-- gil
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