ECE251: Intro to Microprocessors Name: Solutions Mid Term Exam October 4, 2018

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1 ECE251: Intro to Microprocessors Name: Solutions Mid Term Exam October 4, 2018 (PRINT) Instructions: No calculators, books, or cell phones; do not communicate with any other student. One side of a single sheet of 8.5 x 11 inch handwritten notes are allowed. Work all problems, and show all intermediate work, NOT just answers. Return Handout within exam at the end of the exam. Do NOT write in Handout. Extra credit may be given for neatness. 1. (22 pts. total) BASICS Perform the following 8-bit number conversions as requested: 139 = 0x 8B 123 = = (unsigned) 0xBD = (signed) - 19 = 0x ED For the following 32-bit questions, assume N=Z=V=C= 0 before each code sequence. What are the values of r7: 0x C0A N: 0 Z: 0 V: 0 C: 0 following execution of the code segment below: MOV r5, #0xBA5C MOV r6, #0x51AE ADCS r7, r6, r5 What are the values of r4: 0x FABC D321 N: 1 Z: 0 C: 1 following execution of the code segment below: LDR r3, =0xABCD321F ASRS r4, r3, #4 What are the values of r2: 0x FFFF FDFF N: 1 Z: 0 following execution of the code segment below: MOV r0, #0x00007D0F MOV r1, #2_ ; Ignore spaces between bits ORNS r2, r0, r1 A memory subsystem has 9 address pins and 4 data pins. 2 9 x4/8 = 2 8 = 256 Number of bytes of memory in this subsystem.

2 2. (25 pts.) ANALYSIS: For each instruction below, assume that all memory is R/W and: r0 = 0xFF00.A5A5 and r1 = 0x = 0x0100 Show only changes in memory and register values resulting from each instruction. Note that these changes are NOT cumulative; they each execute with the original register and memory values. Examples: STR r0, [r1] r0 = 0x m[0x_0103_:0x 0100 ] = 0x_FF00FEDC LDR r3, [r1] r3 = 0x_CDABB1A0 m[0x :0x ] = 0x a. STR r0, [r1], # -16 r0 = 0x m[0x 0103 :0x 0100 ] = 0x FF00 A5A5 r1 = 0x 00F0 b. LDR r5, [r1, #12] r5 = 0x F5 m[0x :0x ] = 0x r1 = 0x c. STRH r0, [r1, #6]! r0 = 0x m[0x 0107 :0x 0106 ] = 0x A5A5 r1 = 0x 0106 d. STR r0, [r1, # -12]! r0 = 0x m[0x 00F7 :0x 00F4 ] = 0x FF00 A5A5 r1 = 0x 00F4 e. LDRSB r3, [r1, #9]! r3 = 0x FFFF FFC2 m[0x :0x ] = 0x r1 = 0x 0109 Address 0x00F0 0x00F1 0x00F2 0x00F3 0x00F4 0x00F5 0x00F6 0x00F7 0x00F8 0x00F9 0x00FA 0x00FB 0x00FC 0x00FD 0x00FE 0x00FF Data 0xF5 0xE4 0xD3 0xC2 0xB1 0xA0 0x01 0x23 0xAB 0xCD 0x45 0x67 0x89 0xFF 0xEE 0xDD Address 0x0100 0x0101 0x0102 0x0103 0x0104 0x0105 0x0106 0x0107 0x0108 0x0109 0x010A 0x010B 0x010C 0x010D 0x010E 0x010F Data 0xA0 0xB1 0xAB 0xCD 0x45 0x67 0x89 0x01 0x23 0xC2 0xD3 0xE4 0xF5 0x00 0x11 0x22

3 3. (24 pts.) UNDERSTANDING CODE: Your boss, Lucky, would like for you to review her code. Here it is. The comments are correct. Her code may not be: ; This routine analyzes array array. ; The results are stored in register r0 ; Owner: Lucky Last changed: Status: Not completely tested ; This is inserted into SampleCode.s AREA mydata, DATA array SPACE 400 ; assume RAM has 32-bit integer values size EQU 10 AREA mycode, CODE, READONLY, ALIGN=2 EXPORT main main LDR r3, #size ; Put size in r3 ; Initialize r4 and r0 LDR r4, =array ; Load location of array in r4 LDR r0, [r4] ; Load r0 with [r4] (first element) ; Loop through array MOV r2, =1 ; Put 1 in r2 initial counter value loop CMP r2, r3 ; Compare r2 with r3 ckdone BEQ done ; Program done if EQ --go to deadl LDR r5, [r4,#4] ; Load r5 with array element & move pointer CMP r5, r0 ; Compare r5 with r0 test BGE around ; if it s GE, got to around MOV r0, r5 ; Update r0 to value of r5 around ADD r2, r2, #1 ; Add 1 to r2 B loop ; Go back to loop deadl B deadl ; Done. Dead loop END ; Finished with code a. What is this routine trying to do? Tell me about r0. r0 is the smallest (signed) element of array (of type word) b. There are four coding errors in this module. Find two of them. Don t guess. 1. ldr r3, #size ;(=size) 3 ldr r5, [r4, #4] ; (add!). 2. mov r2, =1 ; (#1) 4 BEQ done ; (deadl). c. What value does the assembler assign to the symbol size? 10 or 0xA Smallest value of size for algorithm to work correctly? 1 d. Explain what the line ckdone BEQ done is checking. Checking whether r2 has reached the value of r3 (size) and therefore program has tested all array elements. e. In the instruction labeled test, what is the program s algorithm looking for? To see if the current array entry being tested is the smallest yet found. f. In the instruction LDR r5,[r4,#4], why is #4 used? Since there are 4 bytes in a word, the next word has a 4-byte offset from current word. g. Is data in array signed, unsigned, or either? Signed BGE is a signed number test

4 4. (24 pts.) UNDERSTANDING STACKS AND SUBROUTINES: Consider the following test program and subroutine Smart below. Assume SP is initialized to [SP]=0x : ; Main Program Test Program to exercise subroutine Smart ; AREA mydata, DATA, READWRITE Array SPACE 0x200 ; Starts at 0x ASSUME ALL BYTES=0xA5 Var1 DCW 0x12AB ; Assume [Var1] is initialized to 0x12AB AREA myprog, CODE, READONLY EXPORT main main ; Start of program LDR R1, =Array ; Get parameter address LDR R0, =Var1 ; Get second parameter address LDRB R2, [R0] ; Put the call-by-value data in R2 PUSH {R1,R2} ; Push R2 and then R1 onto stack SUB SP, #4 ; Make ROOM for return data BL Smart ; Call subroutine Smart Callret POP {R7} ; Save return data in R7 ; Undo stack changes, restore registers in main program Deadl B Deadl ; Done ; Smart PUSH {R1} ; Save R1 on stack SUB SP, #8 ; Create local storage for subroutine Smart LDRB R1, [SP, #CBV] ; Fetch call-by-value data from stack ADD R1, R1, #1 ; Increment it STR R1, [SP, #ROOM] ; Put Data where ROOM was made above ; Undo all local storage created in subroutine Smart BX LR ; BX to LR address END ; Finished with program a. What actual code would be used to replace the commented line: ; Undo stack changes, restore registers in main program {assuming Smart actually POPs all its local storage} POP {R1, R2} b. On the next page, fill in the registers and memory table with hex values where memory values are defined, and show (to left of table) where pointer SP points when the processor reaches the location commented with. If a value can t be determined, leave it blank. If it is the address of an instruction, use its label (Deadl, etc.). In addition, to the right side of the memory map show where (i.e. annotate) (1) incoming and return parameters, (2) saved registers, (from any subroutine calls) and (3) local storage (in any subroutines) are stored. c. What should the values of CBV and ROOM be, to make the instructions with operate properly? CBV EQU 0x14 or 20 ROOM EQU 0x0C or 12 d. What is in R0 when program gets to Deadl? [R0] = 0x Assume SP initialized to [SP]=0x

5 4. (Continued) Show SP Memory Value Annotation area: below or Register (Word or Byte) R R R AB SP E8 LR Callret 0x SP initialized to this addr. 0x200005FF 00 0x200005FE 00 0x200005FD 00 0x200005FC AB <- CBV 0x200005FB 20 0x200005FA 00 Array 0x200005F9 02 0x200005F8 00 0x200005F7 00 0x200005F6 00 0x200005F5 00 0x200005F4 AC <- ROOM 0x200005F3 20 0x200005F2 00 Saved 0x200005F1 02 Register 0x200005F0 00 R1 0x200005EF 0x200005EE 0x200005ED 0x200005EC Local 0x200005EB Storage 0x200005EA For 0x200005E9 Smart 0x200005E8 <- SP 0x200005E7

6 EE251 Mid Term Exam 5. (5 pts.) GPIO: The TM4C processor uses memory mapped I/O registers. a. What are two of these registers (precise names not needed). What does each do? 1. E.g. DATA Data input/output register (Many other choices) 2. E.g. DIR Direction (input vs. output) register b. What does PDR in GPIOPDR stand for? Pull Down Resistor c. You have been asked to choose a resistor R in series with an LED running at 0.6 volts, drawing 20 ma. current between ground and +3.6 v. Find R: R = ( )volts/(20x10-3 ) amps = 3000/20 = 150 ohms Honor pledge: I have not given, received, or used any unauthorized assistance on this exam. Signature