(49G) Integer Square Detection
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10-11-2015, 06:48 AM
(This post was last modified: 06-15-2017 01:45 PM by Gene.)
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(49G) Integer Square Detection
The programme ISSQ? economically detects whether an integer is a perfect square, returning 1. for yes, 0. for no.
A problem arises in compiling the programme: the ZINTs in bold type have been processed through the command FPTR2 ^Z>ZH & are in fact hexadecimal ZINTs - the on-calculator compiler attempts to return the four numbers as decimal ZINTs, thus debilitating the programme. Similarly, the decompiler produces the programme as here printed, not indicating the hexadecimal nature of the ZINTs. Anyone have a solution to this compilation problem? ISSQ? :: CK1&Dispatch # FF :: :: 1LAMBIND TRUE ZINT 23; ZINT 11 ZINT 1218:019866014613 ZINT 65 ZINT 402483012450293 ZINT 63 ZINT 202021202030213 ZINT 64 BINT5 ONE_DO :: 1GETLAMSWAP_ FPTR2 ^ZMod FPTR2 ^Z># FPTR2 ^ZBit? SWAPDROP ?SEMI ISTOP-INDEX #2* #1- NDROP FALSE ExitAtLOOP ; LOOP 1GETABND SWAP NOTcsdrpfls :: DUP Z2_ Z< DUP ?SEMI DROPDUP FPTR2 ^Z>ZH FPTR2 ^ZBits BINT2 #/ #+ Z2_ SWAP FPTR2 ^PPow# BEGIN SWAPDROPDUP 3PICKOVER FPTR2 ^ZQUOText FPTR2 ^QAdd Z2_ FPTR2 ^ZQUOText 2DUP Z> NOT_UNTIL DROPDUP FPTR2 ^ZSQ_ ROT EQUAL ; SWAPDROP ; COERCEFLAG ; ; |
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Messages In This Thread |
(49G) Integer Square Detection - Gerald H - 10-11-2015 06:48 AM
RE: HP 49G: Integer Square Detection - Joe Horn - 10-12-2015, 09:41 AM
RE: HP 49G: Integer Square Detection - Gerald H - 10-12-2015, 11:53 AM
RE: HP 49G: Integer Square Detection - Gerald H - 10-12-2015, 12:08 PM
RE: HP 49G: Integer Square Detection - Gerald H - 10-13-2015, 07:04 AM
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