Assembly Language Programming Assignment 1

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Candice Anthony
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1 U08809 Microprocessors Assembly Language Programming Assignment 1 1. Write a short assembly program that illustrates the use of the direct addressing mode, and the use of the MUL function 2 number2 slot Multiply number in a register by b register number2 into arithmetic registers solution from a into total number1 equ 70h ;create simple names for memory slots number2 equ 71h total equ 72h ;address of first instruction mov number1, #06h;put numerical value 6 in slot named number1 mov number2, #02h;and value 2 in 'number2' mov a, number1 ;move number1 into accumulator mov b, number2 ;and number2 into second arithmetic register mul ab ;multiply a by b mov total, a ;move solution into 'total' slot nop ;passive program 1

2 2. Three numbers are stored in memory locations 70h, 71h, and 72h, which should be given the names X, Y and Z. Write an assembly language program that finds the smallest number and stores it in 73h. 2 number2 slot X < Y? Y < Z? Y = Smallest X < Z? Z = Smallest X = Smallest X equ 70h Y equ 71h Z equ 72h smallest equ 73h ;set up named memory mov X, #0A1h ;input numbers mov Y, #0A2h mov Z, #0A3h mov a, X ;put x number in accumulator subb a, Y ;take y from x (a) jc X_is_smallest ;is a =< 0 (x=smallest)? mov a, Y ;put y number in accumulator subb a, Z ;take Z from y? jc Y_is_smallest ; jnc Z_is_smallest ;then z is smallest! 2

3 X_is_smallest: mov a, Z subb a, X jc Z_is_smallest mov smallest, X Y_is_smallest: mov smallest, Y Z_is_smallest: mov smallest, Z finish: nop ;put Z in 'a' ;subtract X ;is a =< 0 (Z=smallest)? ;no? move X into smallest 'variable' ;move Y into smallest variable ;move Z into smallest variable ;done (passive ) 3. Write an assembly language program that finds the mean of three numbers. You can assume that the sum of the three numbers does not exceed FFh 3 number1, 2 & 3 slots value 3 (number of slots) into b register a register value into mean slot Add each number together in accumulator Divide value in a register by b register b register value into remain slot number1 equ 70h ;named memory slots for inputs number2 equ 71h number3 equ 72h mean equ 73h ;slot for mean number (solution) remain equ 74h ;slot for remaining number ;address of first instruction mov number1, #0Bh ;move value into each number slot 3

4 mov number2, #0Ch mov number3, #0Dh mov a, number1 add a, number2 add a, number3 mov b, #03h div ab mov mean, a mov remain, b ;move number1 into accumulator ;add number2 and number3 ;move value 3 into b register ;div register 'a' by register 'b' ;move calculation from a into 'mean' ;move remainder from 'b' register into ;'remain' 4. Use a looping technique to find the sum of 12 numbers stored in locations 60h to 6Bh. All the numbers will be less than 18h. For this particular question I decided to take the question into depth by creating a loop that generates an increasing number in slots 60h to 6Bh, whilst at the same time adding the sum of each slot together in the same looping sequence. This is done using a carried value to increase the number in odd ascing format (e.g. 1,3,5,7...). The sum is then put in its own slot, for good measure, to easily show each slot in action in Keil uvision. value into first slot and put in Add same value to sum ( a register) and carry it directory 0Bh into register 1 Increase slot number for and add carry Increase slot number for and add carry Increase number in new slot by 2 Decrease R1 value by 1 Carry new value and add carry to sum Is R1 = 0? sum in a register to allocated sum slot 4

5 carry equ 6Ch sum equ 6Dh ;to use as carrying number ;to use for final sum mov 60h, #01h ;put numerical value 1 into 1st slot (60h) mov R0, #60h ;move input dir into register 0 (R0) mov R1, #0Bh ;move directory 0B into R1 (loop 11 times) mov ;move value in R0 into accumulator for sum mov ;keep carried number to be used in loop loop:inc R0 ;increment R0 directory by 1 carry ;put carried number in new slot ;increase number in new slot by 2 mov ;move new number into carried slot add a, carry ;add carried value to accumulator for sum djnz R1, loop ;return to loop until R1 directory = 0 mov sum, a ;put sum in it's own box, for good measure nop ;passively the program 5

ET355 Microprocessors Thursday 6:00 pm 10:20 pm

ITT Technical Institute ET355 Microprocessors Thursday 6:00 pm 10:20 pm Unit 4 Chapter 6, pp. 139-174 Chapter 7, pp. 181-188 Unit 4 Objectives Lecture: BCD Programming Examples of the 805x Microprocessor