simulator - LC-3 binary instructions exercise -

this code lc3 simulator have right now:

0011000000000000 0101010010100000 0010011000010000 1111000000100011 0110001011000000 0001100001111100 0000010000001000 1001001001111111 0001001001100001 0001001001000000 0000101000000001 0001010010100001 0001011011100001 0110001011000000 0000111111110110 0010000000000100 0001000000000010 1111000000100001 1111000000100101 0011000100000000 0000000000110000

it's program identitfies single character in entire string , outputs number of times character found..

i have 2 questions..

1) code works outputs number of times find letter in string... need modify builds list in memory of addresses character found.. need make list start @ memory location x3200

2) code works if character found between 0-9 times.. need modify works 0-99 times

i'm not asking answer... appreciate pointers on how approach this...

alright think i'm starting understand problem little bit more. created binary file , loaded lc3's simulator produce readable asm. here's i've got:

memory:  x3000  0101010010100000  x54a0  ,    r2, r2, #0     // clear r2  x3001  0010011000010000  x2610  ld     r3, x3012      // load the value stored @ x3012 r3  x3002  1111000000100011  xf023  trap   in             // input char  x3003  0110001011000000  x62c0  ldr    r1, r3, #0     // load value of memory r3 r1  x3004  0001100001111100  x187c  add    r4, r1, #-4    // add -4 r1 see if have reached end of our string  x3005  0000010000001000  x0408  brz    x300e          // output result, end program if run x0004 in string  x3006  1001001001111111  x927f  not    r1, r1         // invert char   x3007  0001001001100001  x1261  add    r1, r1, #1     // stored string  x3008  0001001001000000  x1240  add    r1, r1, r0     // compare char entered user  x3009  0000101000000001  x0a01  brnp   x300b          // if chars don't match grab char string  x300a  0001010010100001  x14a1  add    r2, r2, #1     // r2 our char counter, counts number of times find                                                        // user's char in string  x300b  0001011011100001  x16e1  add    r3, r3, #1     // increment our memory pointer  x300c  0110001011000000  x62c0  ldr    r1, r3, #0     // load value of memory r3 r1  x300d  0000111111110110  x0ff6  brnzp  x3004          // jump      x3004  x300e  0010000000000100  x2004  ld     r0, x3013      // load value of      x30 r0  x300f  0001000000000010  x1002  add    r0, r0, r2     // add      x30 our count convert ascii  x3010  1111000000100001  xf021  trap   out            // output count user  x3011  1111000000100101  xf025  trap   halt           // stop program  x3012  0011000100000000  x3100  st     r0, x2f13        x3013  0000000000110000  x0030  nop 

to started first question, similar how r3 being used. load memory location x3200 unused register , increment each time store address in memory location. example:

ld r5, x3013      // store x3200 in memory location x3013 . .                 // if chars match following . str r3, r5, #0    // store value of r3 mem of r5 add r5, r5, #1    // increment r5 

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as second question, reason why outputs values 0-9 because you're converting count value ascii char adding x30. great first 10 integers, you'll have little creative add second digit.

hope helps.