2) Using the same instruction set for the TinyProc2, convert the following hex values to assembly language: x0f
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1 CS2 Fall 28 Exam 2 Name: ) The Logisim TinyProc2 has four instructions, each using 8 bits. The instruction format is DR SR SR2 OpCode with OpCodes of for add, for subtract, and for multiply. Load Immediate mode format is DR Immediate4 Registers are r to r3. Convert the following instructions to hex machine code: add r2, r, r3 point each (5 total) mul r, r3, r2 ldi r2, -2 9C 7A BB ldi r, 5 57 add r, r2, r3 6C 2) Using the same instruction set for the TinyProc2, convert the following hex values to assembly language: xf LDI R, 3 2 points each ( total) x53 xbf xc6 xb8 LDI R, 4 LDI R2, - MUL R3, R, R ADD R2, R3, R2 3) If you run the assembly language for the prior question, what value is in r2 when you finish? (decimal)
2 4) You have a finite state machine with 3 states. How many bits do you need to hold the current state? 4 bits 5) Convert the MIPS instruction addi R5, R, -2 into hexadecimal. The op code for add immediate is 8 (6 bits), and RT is the destination register. Op RS RT Immediate (6 bits) xfff4 6 points x265fff4 <- Hex for the instruction. 6) Fill out a truth table for this Finite State Machine (it has three lights, A, B, and C, and an input T that controls the order the lights are shown in). A C B T S S S S A B C x x x x x x x x x x 8 points point per line. If order is scrambled, 4 points off. If total mess, zero points. Blank assumed to be zero for the lights. 7) What kind of map do we use to convert a truth table into a circuit? Starts with a K. Karnaugh
3 8) Your MIPS program executes a JAL instruction at memory location x42a48. Register $s has 7 in it, and the stack pointer is at xa42f84. After the JAL instruction, what value is in register $ra? x42a4c (the instruction location +4) This question has some irrelevant information. 9) Write MIPS or ARM code for the function f int f() f2(); f3(); f: addi $sp, $sp, -4 sw $ra, ($sp) jal f2 jal f3 lw $ra, ($sp) addi $sp, $sp, 4 jr $ra # for saving/restoring RA on stack # point for the jal calls # point for having the jr $ra ) Assume argument functions are passed on the stack. Write assembly code for the following C function. Return the result in register R if you use ARM, or $v if you use MIPS int sum(int a, int b) return a + b; sum: lw $t, ($sp) lw $t, 4($sp) add $v, $t, $t jr $ra # for the load words # point if the add puts the result in v # point for the jr $ra # Many registers could be used; some variation # is fine on this question ) Assume argument functions are passed on the stack, and you have values in $s and $s. Write the MIPS code for this line of C. Hint this is a question about the stack. sum($s, $s); addi $sp, $sp, -4 sw $s, ($sp) addi $sp, $sp, -4 sw $s, ($sp) jal sum addi $sp, $sp, 8 # 4 points for pushing arguments onto stack # 2 points for restoring SP after the call # point for the jal to sum 7 points
4 2) What s in the variable or array X? void f(int a) a = 3; int x; x = 4; f(x+3); printf("x is %d\n", x); void f(int *a) a[] = ; int x[4] = 3, 4, 5, 6; f(&x[2]); // What s in X? void f(int *a) *a = 4; *a = *a + *a; int x; x = 7; f(&x); printf("x is %d\n", x); 4 points 4 points 4 points 4 3, 4, 5, 8 3) In MIPS, the li instruction inserts a value into a register. If the value requires 32 bits, it is broken into an lui, for load upper immediate, and an ori to set the lower 6 bits. Convert the following instruction into the two steps: li $s, xc232fabc 2 points lui $s, xc232 ori $s, $s, xfabc 4) If $s is a pointer to an integer, write MIPS code multiply the number by 6. In C, this would be *x = *x * 6; Four points if the code works, two points if you use a fast way to multiply! 6 points lw $t, ($s) sll $t, $t, 4 sw $t, ($s) # 2 points for the lw/sw steps # 2 points if the number is multiplied by 6 # 2 points (full credit) if there s a shift instead # of a multiply
5 5) Convert the following C code into assembly. You can assume the variables are integers, and just use the MIPS registers. s = ; s = ; while (s!= ) s = s + 4; s = s - ; li $s, li $, loop: beq $s, $, done addi $s, $s, 4 addi $s, $s, - b loop done: points # Check the labels and branches. # if it s close # full credit if it works completely 6) Convert the following C code into assembly. You can assume the variables are integers, and just use the MIPS registers. if (s == s) s = s + s3; else s = s + 4; s = 7; bne $s, $s, elsepart addi $s, $s, $s3 b done elsepart: addi $s, $s, 4 li $s, 7 done points # If/Else sections could be swapped # Looking for bne or beq to branch # Note the b done to skip the else part # if close # full credit if it works Super secret question for one bonus point: what is the greatest rock band of all time? Any answer here is point. Exam total:
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