Paul Gilmartin wrote: <begin extract> It has been patiently explained to me here that STORAGE OBTAIN is biased toward doubleword multiples. But PARM contains a halfword length that doesn't count itself. So I'd expect the limit to be either 32758 (8*4095-2) or 32766 (8*4096-2). Again, why 32760? </end extract>
It is of course possible to quarrel with the rationale, but the derivation is straightforward: q = floor[mod(32767,8)] = 4095, z = 8q = 32760 32760 is the largest integal multiple of 8 not greater than 2^15 - 1 = 32767. Consider now what happens if storage is allocated on a doubleword boundary D. The next doubleword boundary is at D + 8. Placing the halfword current-length prefix at D + 6 ensures that any doubleword alignment for what begins at D + 8 is satisfied. Be grateful that quadword alignment is still only a very specialized requirement. If it were enforced the maximal length of a current-length prefixed PARM string would be 32752. John Gilmore, Ashland, MA 01721 - USA ---------------------------------------------------------------------- For IBM-MAIN subscribe / signoff / archive access instructions, send email to [email protected] with the message: INFO IBM-MAIN
