John [McKown]: The answer to the OP's question depends upon whether the length n under discussion is a length in bytes, as it is and must be in XLn, or a length in bits.
If the former XL2 is sufficient for either +12500 or unsigned 12500. If the latter a fullword will be required because 100000 is greater than either 32767 or 65535. Since, however, the storage is being declared/allocated in bytes, it seems clear to me that the length is/ought to be expressed in bytes too. I do not see much rationale for your different view in this particular context. Note also that your expression for the capacity of a signed halfword is slightly wrong. Because twos-complement representations are used--Zero is its own twos complement--these limits are not symmetrical. They are -2^15 <= H <= +2^15 - 1 -32768 <= H <= +32767. 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
