OK so this:
Quote:
exp: 13*160= 2080
means you need the
inputs to be able to go over 99, not just the output. There is a difference. Of course 9999*9999 is 99,980,001 which needs four bytes for output, and 999*999 is 998001 which needs three bytes of output; but for now I will assume that it is adequate as long as the output is only two bytes, limited to 9999, the same precision as the input, instead of double precision. You cannot have
both inputs be over 99 that way, but I assume that's ok for now.
So we will make both inputs n1 & n2, as well as the output p and "salida" to be two bytes each, assuming there are two bytes available there for each, and change the routine to this:
Code:
salida=io_area+3 ; ("salida" means "output" in Spanish)
salida2= io_area+2
.ORG $0200
sed
STZ p ; Initialize the product as 0.
STZ p+1
LDA n2
BNE f1
LDA n2+1 ; Now we have a high byte to check too.
BEQ fin ; ("fin" means "end" in Spanish)
f1: LDA p
CLC
ADC n1
STA p
LDA p+1 ; Insert these 3 lines for 2nd pair of digits.
ADC n1+1 ; (Do not CLC first this time.)
STA p+1
; DEC n2
LDA n2
SEC
SBC #1
STA n2
BNE f1
LDA n2+1 ; If the low byte turned to 0, check the high byte.
BEQ fin ; If it was already 0, you're done.
SBC #1 ; Otherwise decrement it
STA n2+1
BRA f1 ; and continue looping.
fin: LDA p ; Now transfer both output bytes to "salida".
STA salida
LDA p+1
STA salida+1
BRK
n1: .db $10 ; multiplicand
n2: .db $15 ; multiplier
p: .DB $0 ; product
For getting into numbers that big however, adding n2 times is not efficient timewise. Usually the computer is made to internally keep and handle numbers in hex and only convert to and from decimal when it's time for human I/O and the multiplication is done by shifting bits and adding with a much shorter process. There are several hex multiplication routines at
http://6502.org/source/ and
http://6502org.wikidot.com/software-math-intmul and
http://6502org.wikidot.com/errata-softw ... forth#toc1
I did not try the code above, but it's simple enough that hopefully I didn't make a dumb mistake.
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