202201-midterm-1

Page 3 of 10 Question 4: The decimal number -27 represented as a 10-bit two’s complement number is equivalent to: _____ 0b1000011011 _____ 0b1111100100 __X__ 0b1111100101 _____ 0b1000000101 _____ None of the above. Question 5: The six-bit two’s complement number 0b101111 is equivalent to: _____ 0b00101111 as an eight-bit two’s complement number. _____ -31 _____ -15 _____ -29 __X__ None of the above. Question 6: If we consider I/O port B, then: _____ we set the data-direction for each bit (input or output) by an appropriate value stored in the PINB and PORTB registers. __X__ we set the data-direction for each bit (input or output) by an appropriate value stored in the DDRB register. __X__ we read values from the port by reading from the PINB register. _____ we read values form the port by reading from the PORTB register. _____ None of the above.

Page 6 of 10 Section B (20 marks): One question Question 13: Consider the following program written in AVR assembler: .cseg .org 0 start: ldi r16, 0b00001010 ldi r17, 0b00000101 pointA: add r16, r17 dec r18 dec r17 cpi r17, 0x02 brne pointA stop: rjmp stop a) How many executions occur for the dec r17 instruction? Explain your answer. It executes three times, such that the value in r17 goes from 5 to 4, then from 4 to 3, and from 3 to 2. At this third decrement, the value in r17 is equal to 2, which means the cpi instruction will results in the Z flag being set, and brne will not track to pointA , such that the program proceeds to the rjmp loop. (8 marks) b) What is the value in r16 when the program reaches the instruction rjmp stop ? Explain your answer. This code fragment repeatedly adds the value of r17 to r16 , although r17 is decremented by 1 each time through the loop: 1. First add: 10 + 5 = 15 2. Second add: 15 + 4 = 19 3. Third add: 19 + 3 = 22 Therefore the final value stored in r16 is 22 ( 0x16 ) (6 marks)

Page 9 of 10 Question 15: 10 marks Consider the following encoding of an instruction: 0xF7A4 Using the information provided to you in the “AVR Architecture Reference Booklet” that comes with this exam, what is the actual AVR instruction (i.e. mnemonic plus argument)? Your answer must not only show all work but also explain how you arrived at your answer. The binary of the instruction (separated into groups of four bits) is as follows: 1111 0111 1010 0100 Examining the 16-bit opcodes provided in the AVR Architecture Reference Booklet provided with the test, there is a single instruction that begins with 1111 : the BRGE instruction. This instruction has the following encoding for the value of k (which is a 7-bit two’s complement number, representing values from -64 to +63): 1111 01KK KKKK K100 Therefore the value of KK KKKK K here is 11 1010 0 . That is, 1110100 is a negative number (which you can immediately tell because of the left-most set bit). Negating all of this in the conversion technique shown in the course produces 0001100 – which is 12. Therefore the value of k is -12. Putting this all together, the instruction is: BRGE -12 • Good answer: 10 marks • Answer is incorrect, but due to minor mistake: 7 marks • Answer is incorrect, but due to significant mistake: 4 or 5 marks • Failing answer (various): < 4 marks