In a message dated 11/8/2007 10:02:17 A.M. Central Standard Time, [EMAIL PROTECTED] writes:
I have not checked this for myself yet, and probably won't have the time in the next few weeks... In theory... If I allocate 200 bytes of storage at x'7fffff00', and my program, the way some Cobol programs are, writes 1600 bytes there, what would the addresses be, would it jump across the 'hole' or try and write over the hole? The addresses of the 200 bytes (decimal number 200 is assumed) that your program allocates would be X'7FFFFF00' to X'7FFFFFC7'. If your COBOL program attempts to write 1600 (decimal assumed again) bytes beginning at X'7FFFFF00', your program would write at most the first 256 of the 1600 bytes and, when attempting to write the 257th byte, would be interrupted with a protection check program interrupt that would result in a S0C4 ABEND. I said "at most" because I have not yet studied the latest PoOps to know how new move instructions move bytes. And it would depend on how your program does the move. If it moves one byte per instruction in a loop, it would write 256 bytes and then ABEND. There might be another type of move instruction whose preprocessing checks the beginning and ending addresses for validity before moving the first byte, in which case your program would write zero bytes and then ABEND. Your program would try to write over the hole but would not be allowed to by the various protection mechanisms in the processor architecture. Bill Fairchild Franklin, TN ************************************** See what's new at http://www.aol.com ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to [EMAIL PROTECTED] with the message: GET IBM-MAIN INFO Search the archives at http://bama.ua.edu/archives/ibm-main.html
