Assembly Language Help

Question

I know this proably not the right place to post this topic but I couldn't find anywhere else.
Basically I am developing a program for high school kids learning assembly language. I am developing it in Java but it has been over 10 years since I have ever done any assembly. I am working through good books and tutorials to refresh my mind. I have come to the example below and got stuck. can anyone spare 5 mins to explain ech line of the following to me in real simple terms. Duke dollars available.
Load R1,R8
add R1,16
Loadind R2,R1
add R1,8
loadind R3,R1
add R3,R2
sub R1,4
storeind R3,R1
Many thanks

Answer 1

It looks like an addition of operands on the heap of a 32bit processor.
R8 is a pointer to a heap which has the following format:   +-------+-------+-------+-------+-------+-------+-------+
   |       |       |       |       |   A   |   C   |   B   |
   +-------+-------+-------+-------+-------+-------+-------+
   R8    R8+4    R8+8    R8+12   R8+16   R8+20   R8+24What we want is to execute C=A+B.
Load R1,R8We are going to have R1 pointing on the place in the heap we are interested in. R8 is probably a register reserved for the heap, which should not be changed, so we copy it into R1.
add R1,16Move R1 pointer so that it points 16 bytes higher, where A is.
Loadind R2,R1Load the content of A (pointed by R1) into R2. Note that we are not interested here in the value of R1, but in the value pointed by R1, thus the "ind" for "indirect".
add R1,8Move the R1 pointer to point 8 bytes higher, where B is.
loadind R3,R1Load the content of B (pointed bt R1) into R3.
add R3,R2Add R2 to R3 (the result is stored into R3).
sub R1,4Move the R1 pointer to point 4 bytes lower, where C is.
storeind R3,R1Store the result into C.
Jerome.