Assembly at&t syntax, return an ascii integer

Question

This code below takes in an two digit number (let's say 12) in ASCII and should return this number as the number 12. (NOT ASCII)

mov     $IBUFF, %eax    # IBUFF stores the ASCII number 12
movl    $NUMBER, %ecx   # NUMBER will i put the number 12 (NO ASCII)

movb    (%eax), %bl     # Moves the first byte (1) to %bl
subb    $48, %bl        # Decrease %bl so it is not longer a ASCII sign.
movb    %bl, (%ecx)     # Moves this byte into the first place in NUMBER

inc %eax                # Increase IBUFF so it points to the next integer 
inc %ecx                # Increase NUMBER so it points to the next empty space

movb    (%eax), %bl     # Same as above
subb    $48, %bl        # Same as above
movb    %bl, (%ecx)     # Same as above


mov NUMBER, %eax        # Move the number 12 from NUMBER to %eax (return register)

When i run this code the return value is 513, not 12. What am i doing wrong?

Ori Pessach · Accepted Answer · 2012-02-09 14:22:09Z

1

You need to accumulate the result (possibly in a register) by multiplying it by 10, and then adding the new digit. Basic base 10 representation. What you're doing ends up storing the resulting digits in base 256.

answered Feb 9, 2012 at 14:22

Ori Pessach

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Ori Pessach Over a year ago

@Bewn - No, but multiplying a number by 10 and then adding another number to it shouldn't be that hard...

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