Question

Please help by showing steps.

Question 4. (continued) (b) Consider a 16-bit binary number stored in AVR registers r15:r14 which the programmer considers to be a two's complement value. (r15 holds the most significant byte, r14 holds the least significant.) Write down a sequence of AVR assembly language instructions which perform each of the following operations The result should end up in r15:r14. Other registers can be used freely if required i) Sets r15:r14 to the constant value-1 (2 marks) Ser r16 Mov r15, r16 Mov r14,r16 ii) Multiplication of r15:r14 by 5 Mov r17, r15 (2 marks) Mov r16, r14 Lsl r14 Rol r15 Lsl r14 Rol r15 Add r14,r16 Adc r15,r17

Answer 1

//i am asuming you are willing to find out how the 2 programs work
1.

    Ser r16
    Mov r15,r16
    Mov r14,r16

explaination:
    here first instruction is Ser ie set all bits
    setting all bits means 1111111111111111
    which is in 2's compliment form
    equivalent decima number of this is -1
    so we are loading -1 in r16

    2nd and 3rd instructions are Mov, here 1st register is destination and 2nd is source
    hence we are putting -1 in both registers r14 and r15

2.

    Mov r17,r15
    Mov r16,r14
    Lsl r14
    Rol r15
    Lsl r14
    Rol r15
    Add r14,r16
    Add r15,r17

explaination:
    1st and 2nd are Mov instructions as described above, here we are loading r17 with r15 and r16 with r14
  
    3rd instruction is Lsl, lets see what it does:
    Shifts all bits in R14 one place to the left. Bit 0 is cleared. Bit 7 is loaded into the C flag of the status register
    This operation effectively multiplies r14 by two.

    next instruction is Rol which is Rotate Left trough Carry, let's see what it does:
    Shifts all bits in r15 one place to the left. The C flag is shifted into bit 0 of Rd. Bit 7 is shifted into the C flag

    as this is mentioned that r15 is most significant byte so what we are doing with Rol is multipliying it by two with
    carry generated through Lsl

    so till now we have multiplied r14:r15 with two

    next 2 instructions are again Lsl and Rol which again multiplies r14:15 by two

    so till now we have multiplied r14:r15 with four

    next two instructions are Add instructions where we are adding r16 to r14 and r17 to r15

    equivalent high level code:
                                                r17=r15
                                                r16=r14
                                                r14=2*r14
                                                r15=2*r15
                                                r14=2*r14
                                                r15=2*r15
                                                r15=r15+r17
                                                r14=r14+r16

    so finally we have multiplied r14:r15 by 5

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