Question

Assume the memory display of the HCS12 shows 16 bytes starting at S0800 as follows SOBO0 80 53 05 28 36 89 00 FF FE 80 91 3E 77 AB 8F 7F Cavt the results of the following instructions SP S0805 pula A ?, SP SP S08OAL pula pshb A 7, SP -? SP S0805 pula A ? pulb B Idaa $O803 Idab $0804 aba A 7,8- SP $0805 psha pshb SP 7 Idx S0800 idab $0807 leax B.X X 6-What does the following assembly program do? Describe the result of executing the program; don't just say what each instruction is doing. Instructions Idx aS0800 Idaa #5 Idab #255 stab 0x loop incb inx dbne a,loop 7- What is the time delay (in ms) that the following loop generates? Assume a 25 MHz bus clock Consider that pusha takes 2 cycles, pula 3 cycles, nop one cycle and dbne 3 cycles Idx a56000 ; 1 cycle loop: nop ; 1 cycle ; 3 cycles ; 2 cycles ; 3 cycles 2 cycles nop pula psha pula psha pula ;3 cycles dbne X, loop ; 3 cycles Page 11 of 12

Answer 1

Answer :- 5) In HCS12, when a byte is pushed in to the stack, the stack pointer value decreases first by 1 and the value is kept at that memory.

When a byte value is pulled, the current SP memory value is returned ans SP = SP + 1.

The value of registers are shown below-

Assume the memory display of the HCS12 shows 16 bytes starting at 50800 as follows SOBO0: 80 53 05 28 36 89 00 FF FE 80 91 3E 77 AB 8F 7F Gve the results of the following instructions SP S0805 pula A 7, SP SP S08OA pula pshb A 7, SP ? A=$89, SP = $0806 A= $91, SP = $080A SP S0805 pula A pulb B Idaa $0803 Idab $0804 aba A 7, 8- A=$89, B = $00 SP = $0807 A = $2B, B = $36 after ABA, A = A+B = $61 SP $0805 psha pshb SP d S0800 Idab $0807 leax 8,X X- X = $0800, B = $FF after LEAX, X= X+B= $08FF SP = $0803 ne the proeram: don't

Answer :- 6) The given program stores the value in B register to the memory address $0800. This is done for 5 times. The count value 5 is present in register A. After storing the value-

1. The memory address present in X register is incremented by one.
2. The value present in B register is incremented by one.
3. The count value in A is decremented by 1 and if this is not zero then again the value in B register is copied to next memory address in X register.

Thus five bytes of data is being copied to memory. The memory and data values are-

$0800 = 0xFF = 255, count value in A = 5
$0801 = 0x00 = 0, count value in A = 4
$0802 = 0x01 = 1, count value in A = 3
$0803 = 0x02 = 2, count value in A = 2
$0804 = 0x03 = 3, count value in A = 1

When A = 0, the loop ends and program is finished.