I need to make a program how pick two number between 0 and 9 and multiply they but I can only use SHL, SHR or ROR,ROL commands

;I find this code here on stackoverflow but a don´t really undusted 
    ; ax = x
    mov bl, al     ; bx = x
    shl bl, 3      ; bx = 8 * x
    add al, bl     ; ax = 9 * x
    shl bl, 2      ; bx = 32 * x
    add al, bl     ; ax = 41 * x

CodePudding user response：

Performing multiplication without using the mul instruction is clear enough, but saying you can only use shl, shr, rol, and ror instructions is not sufficient to solve this task.
As an example see the below subroutine that manages to multiply without the use of the mul instruction:

; IN (al=first number [0,9], ah=second number [0,9]) OUT (ax=product [0,81])
SmallMul:
  mov [$ 6], al ; This overwrites the default operand of AAD
  mov al, 0
  aad          ; -> AX is product
  ret

and using add and sub it is possible to make this program ?

Yes, using repeated addition

example: 5 * 3  -->  0   5   5   5
                      <--- 3x --->

; IN (al=first number [0,9], bl=second number [0,9]) OUT (cl=product [0,81])
SmallMul:
    sub  cl, cl  ; Sets CL=0
.a: add  cl, al
    sub  bl, 1
    jnz  .a      ; Jumps BL-1 times
    ret

• If ever the second number in BL were zero, then this simple code will fail!
• As an optimization we can choose the smallest of the two numbers to control the loop. Less iterations is a good thing.

; IN (al=first number [0,9], bl=second number [0,9]) OUT (cl=product [0,81])
SmallMul:
    sub  cl, cl  ; Sets CL=0
    cmp  bl, al
    jbe  .b
    xchg bl, al
    jmp  .b
.a: add  cl, al
.b: sub  bl, 1
    jns  .a      ; Jumps BL times
    ret

A simple version of the shift and add algorithm:

; IN (al=first number [0,9], bl=second number [0,9]) OUT (cl=product [0,81])
SmallMul:
    sub  cl, cl  ; Sets CL=0
    jmp  .c
.a: add  cl, al
.b: shl  al, 1   ; Doubles AL, same as `ADD AL, AL`
.c: shr  bl, 1   ; Halves BL
    jc   .a      ; Shifted out a 1 bit, so go ADD
    jnz  .b      ; Jumps for as long as BL has non-zero bits
    ret

• There's no problem with the second number in BL being zero.
• As an optimization we can choose the smallest of the two numbers to control the loop. Less iterations is a good thing.

; IN (al=first number [0,9], bl=second number [0,9]) OUT (cl=product [0,81])
SmallMul:
    sub  cl, cl  ; Sets CL=0
    cmp  bl, al
    jbe  .c
    xchg bl, al
    jmp  .c
.a: add  cl, al
.b: shl  al, 1   ; Doubles AL, same as `ADD AL, AL`
.c: shr  bl, 1   ; Halves BL
    jc   .a      ; Shifted out a 1 bit, so go ADD
    jnz  .b      ; Jumps for as long as BL has non-zero bits
    ret