如何在MC68k中将ASCII码转换为二进制

Question

I have to write a program that requires 20 user inputted numbers from 0-100 to find the average of the numbers and categorize them as failing or passing but it saves the input as ascii in memory and I have to change it from ascii to binary. I know that ascii numbers are 30-39 in hex but I am unsure about how to implement it in MC68K like if i input 93 as a number then it would be saved as 3933 but how do I convert it to binary?

Answer 1

  clear number
loop:
  get highest digit character not dealt with yet
  compare digit character to '0' ; that's character 0, not value 0
  blo done
  compare digit character to '9'
  bhi done
  multiply number by 0x0a ; 10 in decimal
  subtract 0x30 from the character
  add to number
  jmp loop
cone:
  ...

Answer 2

Here's how I would do it:

str_to_int:
    ; Converts a decimal signed 0-terminated string in a0 into an integer in d0.
    ; Stops on the first non-decimal character. Does not handle overflow.
    ; Trashes other registers with glee.
    moveq #0, d3
.signed:
    cmpi.b #'-',(a0)     ; Check for leading '-'.
    bne.s  .convert
    bchg   #0,d3
    addq.l #1,a0
    bra.s  .signed
.convert:
    moveq.l #0,d0
    moveq.l #0,d1
.digit:
    move.v (a0)+,d1
    beq.s  .done
    subi.b #'0',d1       ; Convert to integer.
    bmi.s  .done         ; If < 0, digit wasn't valid.
    cmpi.b #'9'-'0',d1
    bgt.s  .done         ; If larger than 9, done.
    muls.l #10,d0
    add.l  d1,d0
    bra.s  .digit
.done:
    tst.b  d3
    bne.s  .signed
    rts
.signed:
    neg.l  d0
    rts

Note: the above is untested assembly code for a processor I haven't touched in ... quite a while. Hopefully it can at least be inspirational. Back in the day, of course nobody would have dared use muls here, but it's instructional. Pardon the pun.