.data
arr1 DB 4 dup (0),9
arr2 DB 4 dup (0),9
arr3 DB 5 dup (0)notice that, now, the 9's are the 5th element in arrays 1 and 2
since the days of the 8008, intel has provided special instructions to help handle simple BCD math
AAA ASCII adjust after addition
AAS ASCII adjust after subtraction
AAM ASCII adjust after multiplication
AAD ASCII adjust before division
DAA decimal adjust after addition
DAS decimal adjust after subtraction
even though the "ASCII" names imply the numbers are ASCII, they aren't :P
they are unpacked BCD numbers that may easily be converted to ASCII when done
these instructions generally work on 2 digits at a time, one in AL, one in AH
the "decimal" instructions are used with packed BCD values
packed BCD has 2 digits in each byte, one in the lower 4 bits, one in the upper 4 bits
there are no DAM or DAD instructions
so, to answer the addition question
mov al,arr1[4] ;the 5th element of array 1
mov ah,0
add al,arr2[4]
aaathe "ones" digit is in AL, the "tens" digit is in AH
there is a shortcut you can use to load bytes and clear higher-order bytes in a register
movzx eax,byte ptr arr1[4] ;zero-extend a byte into EAX
add al,add2[4]
aaa
the AAA instruction also works for ADC
you can replace the array indexes with a register that holds a variable index
mov edx,4
movzx eax,byte ptr arr1[edx]
add al,arr2[edx]
aaa
you can create a loop, you just have to move AH into AL after storing each result digit, and adding it into the next digit
you can use google to find descriptions of all the BCD instructions