1. Number Conversions [8 marks]

Transcription

1 1. Number Conversions [8 marks] a. convert the decimal number 42 to a 16-bit signed integer in binary 0b b. convert the 8-bit number 0x42 to a 16-bit signed integer in binary 0b c. convert the signed 8-bit number 0x5A to a 32-bit signed integer in hexadecimal 0x A d. convert the signed 7-bit number 0x5A to a 32-bit signed integer in hexadecimal 0xFFFFFFDA e. convert -2^16 to a 32-bit signed integer in hexadecimal 0xFFFF0000 f. encode 2.25 in 32-bit floating point (IEEE 754), show the result in binary. Show how each of the parts of the floating point number is found. Sign: + => 0 Integer part: 2 = 0b010 Decimal part:.25 = 0b Mantissa = 0b thus we use 0b in the representation Exponent = 1 thus we use =128 = in the representation Answer is thus 0b Page 1 of 9 Midterm ECE243 March 2016

2 Instructions [14 marks] Operations [8 marks] r2 0x r3 r4 r5 0x x xF0F0F0F0 Example solutions: movi r2,0 ans: r2 = 0x Address 0x1000 0x1004 0x1008 Data Word 0x x x stw r0,0(r0) ans: mem[0] = 0x Indicate the new 32-bit value and register or memory location that changes after executing the following instructions, using the format from the example. Use the same initial state above for each part. Note that memory is shown as a set of words, so you do not need to consider endianness in this question. a) add r4, r3, r2 r4=0x c b) slli r4, r3, 4 r4=0x c) ldw r3, 4(r2) r3=0x d) stw r3, 0(r2) mem[1004]=0x e) mov r4, r3 r4=0x f) movi r2, -1 r2=0xffffffff g) and r2, r2, r5 r2=0x h) or r2, r2, r5 r2=0xf0f0f0f4 Page 2 of 9 Midterm ECE243 March 2016

3 2.2 Sign extension and endianess [6 marks] Show the result of the last instruction in each sequence for Nios II (little endian) given the following initial state of memory. Sometimes it might be changes in memory; sometimes it might be changes in register contents. Use the same initial state for each part. Use the notation rx = 0x######## for register changes or mem[address] = 0x## for each byte in memory that gets updated (you may need multiple lines). Example: mov r2,r0 Answer: mem[0] = 0x00 stw r2,0(r0) mem[1] = 0x00 mem[2] = 0x00 mem[3] = 0x00 Address 0x0 0x1 Data Byte 0x0a 0x1b c) movi r2, 4 ldh r3, 2(r2) r3=0xffffd7c6 0x2 0x3 0x4 0x5 0x6 0x7 0x2c 0x3d 0xa4 0xb5 0xc6 0xd7 a) movi r2, 0 ldw r3, 0(r2) r3=0x3d2c1b0a b) movi r2, 4 ldhu r3, 0(r2) d) movi r2, 2 ldbu r3, 3(r2) r3=0x000000b5 e) movi r2, 3 ldb r3, 0(r2) r3=0x d f) movi r2, 2 movi r3, -2 sth r3, 0(r2) mem[0x02]=0xfe mem[0x03]=0xff r3=0x0000b5a4 Page 3 of 9 Midterm ECE243 March 2016

4 3. Calling conventions [6 marks] You are writing an assembly subroutine foo, which must interface with functions written in C: foo is called by main, and foo calls bar1 and bar2. All of the code for foo is provided below. For each push and pop, circle only the registers that need to be saved (pushed) and restored (popped). Recall caller-saved and callee-saved registers. Use as few pushes and pops as possible maintaining convention: Do not save any registers that you know will not be overwritten or will not be read later; you will lose marks otherwise. foo: push sp, ra, r2, r4, r5, r12, r13, r14, r15, r16, r17 movia r12, 0x10000 # memory address 1 movia r13, 0x20000 # memory address 2 movi r14, 0 # accumulator; return value movi r15, 0xFF # bit mask 1 movi r16, 0xF # bit mask 2 movi r17, 3 # scaling factor and r4, r4, r15 ldw r5, 0(r12) Example If you think r2, r5 and r15 need to be saved, you would circle those like so: push sp, ra, r2, r4, r5, r12, r13, r14, r15, r16, r17 push sp, ra, r2, r4, r5, r12, r13, r14, r15, r16, r17 call bar1 # arguments r4 and r5; returns r2 pop sp, ra, r2, r4, r5, r12, r13, r14, r15, r16, r17 stw r2, 4(r12) add r14, r14, r2 and r4, r4, r16 ldw r5, 0(r13) push sp, ra, r2, r4, r5, r12, r13, r14, r15, r16, r17 call bar2 # arguments r4 and r5; returns r2 pop sp, ra, r2, r4, r5, r12, r13, r14, r15, r16, r17 stw r2, 4(r13) add r14, r14, r2 mul r14, r14, r17 and r14, r14, r16 mov r2, r14 pop sp, ra, r2, r4, r5, r12, r13, r14, r15, r16, r17 ret Page 4 of 9 Midterm ECE243 March 2016

5 4. Understanding assembly [11 marks] a. [5 marks] Write the equivalent C code for this subroutine. Note that you may be able to do parts b and/or c even if you don t get this part. foo: bar: bne r0, r4, bar movi r2, 1 ret # part c addi sp, sp, -8 stw r4, 0(sp) stw ra, 4(sp) addi r4, r4, -1 call foo ldw r4, 0(sp) ldw ra, 4(sp) mul r2, r2, r4 addi sp, sp, 8 ret integer foo(integer n) { if (n == 0) return 1; else return n*foo(n-1); } b. [1 mark] What well-known function does this subroutine implement? n! or n factorial Page 5 of 9 Midterm ECE243 March 2016

6 c. [5 marks] Assume the subroutine on the last page is called like this: main: movia sp, 0x movi r4, 4 call foo Draw the contents of the stack (addresses and word data) used by foo and its descendants immediately before executing the ret instruction with the comment part c (the third instruction of the subroutine). Also indicate the current value of the stack pointer. Assume main = 0x1000 and foo = 0x2000. sp value = 03FFFFE0 Memory Address (words) Data 03FFFFE0 1 03FFFFE4 0x FFFFE8 2 03FFFEC 0x FFFFF0 3 03FFFFF4 0x FFFFFF8 4 03FFFFFC 0x Page 6 of 9 Midterm ECE243 March 2016

7 5. New device [12 marks] Below is a new device for the Nios II, a UART serial device, but different than the ones you have been using. For this one you need to set the bit rate, the data size for each transmission (in bits) and the number of stop bits, and you can set parity. There is no FIFO with the input or with the output, only simple singlemessage buffers. Note: bit rate as used here is equivalent to baud rate or bits of information over the wire (including all the extra bits that surround the actual message) Produce commented code that does the following. Assume the processor clock rate is 1 MHz. a. i) Set the bit rate to 2000 bits per second ii) Start the receiver and transmitter with even parity, normal number of stop bits, 8 bits per message (data sent/received) and with interrupts enabled # 2000 bits/sec => /2000 = 500 bits/sec movia r10,0xff2f0000 #set base address (used in all parts of the answer) movi r11,500 #set baud rate to 500 clocks/bit stwio r11,8(r10) #end of answer to i part movi r11,0b #set control bits for operation as requested stwio r11,12(r10) #end of answer to ii part Page 7 of 9 Midterm ECE243 March 2016

8 b. Write the additional lines of code that would check to see if a character had been input, and that will read the character if into r10 if it was ready. Assume the initialization specified in part a, except with interrupts disabled. You need only write the lines that actually answer this question and not any context, such as putting it into a subroutine. CheckRead: #assume r10 still set to base address ldw r11,4(r10) #get status information andi r11,r11,0x01 #check read bit (note: could check for errors here, wasn t asked for) beq CheckReadDone #br if no character available ldw r11,0(r10) #get read character # one would process the read character in r11 here CheckReadDone: #done the read check c. Write the initialization code that, in addition to the code in part a, would be required to make interrupts active for this device on this processor. Assume there are no other interrupting devices. The interrupt should be fully active on the processor after the code from part a and the code from part c are executed. #there are 2 more steps to do, since the interrupts at the device were already enabled movi r11,0x0400 #set interrupt enable #10 wrctl ienable,r11 movi r11,0x01 #set the PIE bit for global interrupt enable wrctl status,r11 Page 8 of 9 Midterm ECE243 March 2016

9 6. Write a subroutine [18 marks] You have a linked list with each node implemented as follows: struct node { int data; struct node* next; }; The following function allows the user to insert a new node after a node of given index (head node has an index of 0) in the linked list. The insertion function is implemented as follows: void insert(int insert_after, struct node *newnode, struct node *head)//the data for newnode is already set { struct node *prev = head; // You can assume insert_after will always lie within the linked list for (int index = 0; index < insert_after; index++) { prev = prev->next; } struct node *before = prev; struct node *after = prev->next; newnode->next = after; before->next = newnode; } return; You are to write the insertion function in Nios II assembly language. Be sure to follow the calling convention of the Nios II ABI. #note: on call r4=insert_after (# of nodes), r5=*newnode, r6=*head insert: beq r4,r0,atinsertposition addi r4,r4,-1 ldw r6,4(r6) #go to next node br insert #continue until at insertion point atinsertposition: ldw r4,4(r6) #pick up pointer stw r5,4(r6) #replace with pointer to new element stw r4,4(r5) #point new element with next in line ret #done # note that this used parameter registers to avoid having to save registers on the stack. # any callee-save registers used must be saved on the stack Page 9 of 9 Midterm ECE243 March 2016