Difference between revisions of "Square roots in 6502 machine code"
(Created page with "Category:6502 A simple way of calculating integer square roots is to count how many times increasing odd numbers can be subtracted from the starting value. For example: <...")
Revision as of 00:09, 8 March 2015
A simple way of calculating integer square roots is to count how many times increasing odd numbers can be subtracted from the starting value. For example:
32-1=31 -> 31-3=28 -> 28-5=23 -> 23-7=16 -> 16-9=7 -> 7-11<0 1 2 3 4 5 -> SQR(30)=5, remainder 7
This can be done in 6502 machine code as follows:
\ Calculate 16-bit square root \ ---------------------------- \ On entry, (in,in+1)=input value \ On exit, Y=(out)=root, X=(out+1)=remainder \ If out=in+2 this lets you use !in=value .sqr :\ On entry, !in=input value LDY #1:STY out+0:DEY:STY out+1 :\ Initialise out to first subtrand .sqr_loop :\ Repeatedly subtract increasing SEC :\ odd numbers until in<0 LDA in+0:TAX:SBC out+0:STA in+0 :\ in=in-subtrand, remainder in X LDA in+1:SBC out+1:STA in+1 BCC sqr_done :\ in<0, all done INY :\ LDA out+0:ADC #1:STA out+0 :\ step +2 to next odd number BCC sqr_loop :\ no overflow, subtract again INC out+1:BNE sqr_loop :\ INC high byte and subtract again .sqr_done :\ Y=root, X=remainder STY out+0:STX out+1 :\ out?0=root, out?1=remainder RTS
You can test this with:
FOR A%=1 to 65535:!in=A%:CALL sqr:P.A%,out?0,out?1:NEXT
FOR A%=1 to 65535:!in=A%:B%=USR sqr:P.A%,(B%AND&FF00)DIV256,(B%AND&FF0000)DIV65536:NEXT
It is simple to reduce the code to calculate roots of 8-bit numbers by removing the code to subtract in+1. The code uses an 8-bit counter for the root, so to calculate roots of numbers large than 16 bits different code is needed, as the square root of &10000 (a 17-bit number) is &100 (a 9-bit number).
Jgharston 21:37, 30 August 2007 (BST)