CSCI 250 Computer Organization and Assembly Language Programming Spring 2013 G. Pothering Assignment 3 – Solutions Note: No final answer, even if correct, will receive any credit (that is, credit = 0) unless there is an appropriate amount of work shown that leads to the answer. 1. (0 points) Page 146 problem 5.9 – I should not have included this, and so will not count it. 2. (3 points) Page 146, problem 5.23: Suppose the following LC-3 program is loaded into memory starting at location x30FF. If the program is executed, what is the value in R2 at the end of execution? 1110 0110 1111 0001 0001 0010 0100 0000 0100 0100 0000 0100 0010 0100 1000 0001 0010 0101 0001 0010 = = = = = 1110 0110 1111 0001 0001 001 0 0000 0001 010 001 000010 0000 0010 0101 010 001 000 001 010 010 000 010 = = = = = LEA R1 x001 LDR R2, R1, #2 TRAP x25 ADD R2, R1, R1 ADD R2, R2, R2 Results of program execution x30FF LEA R1, x001 x3100 LDR R2, R1, x2 x3101 TRAP x25 x3102 ADD R2, R1, R1 x3103 ADD R2, R2, R2 R1 <-- x3100 +x0001 = x3101 R2 <-- value in memory location (R1 = x3101 + x2) = value at x3103 = x1482 HALT Note, for those who did not stop at the HALT instruction, but continued on to the last "instruction" the results would be the following x30FF LEA R1, x001 x3100 LDR R2, R1, x2 x3101 TRAP x25 x3102 ADD R2, R1, R1 x3103 ADD R2, R2, R2 3. R1 <-- x3100 R2 <-- value value HALT R2 <-- x3101 R2 <-- x6202 +x0001 = x3101 in memory location (R1 = x3101 + x2) = at x3103 = x1482 + x3101 = x6202 + x6202 = xC404 (4 points) Page 146, problem 5.14. Give your answer in hex. Recall DeMorgan's law: NOT (x AND y ) = NOT x OR NOT y The LC-3 does not have an OR instruction; however we can write a sequence of instructions to implement the OR operation. Fill in the two missing instructions -- (2) and (4) -- so the instruction sequence performs OR R3, R1, R2 Solution: By DeMorgan's Law NOT(x AND y) = (NOT x) OR (NOT y), whence NOT ((NOT x) AND (NOT y)) = NOT(NOT x) AND NOT(NOT y) = x OR y We use this in implementing the solution below (1) (2) (3) (4) 1001 1001 0101 1001 100 101 110 011 001 010 100 110 111111 111111 000 101 111111 NOT NOT AND NOT R4, R5, R6, R3, R1 R2 R4, R5 R6 Translate the following high-level statements into LC-3 mnemonic machine language. In all cases assume that: variables i and j are integer variables associated with the registers R0 and R1 respectively the elements of the array A are allocated sequentially in memory starting at address x3018. the first instruction of your code is stored in memory at address x3000. 4. (2 points) A[2] = j x3000 x3001 LEA R2 x017 STR R1, R2, #2 R2 contains address x3001 + x017 = x3018 5. (3 points) j = A[2] AND A[3] x3000 x3001 x3002 x3003 LEA LDR LDR AND R2 x017 R3, R2, #2 R4, R2, #3 R1, R3, R4 R2 contains address x3001 + x017 = x3018 6. (2 points) j = A[2] - 1 x3000 x3001 x3002 LEA R2 x017 LDR R4, R2, #2 ADD R1, R4 #-1 R2 contains address x3001 + x017 = x3018 7. (3 points) j = A[i-1] x3000 x3001 x3002 LEA R2 x017 ADD R4, R2, R3 LDR R1, R4, #-1 R2 contains address x3001 + x017 = x3018 R4 contains address A + i R1 gets the value at address (A+i) - 1 8. (3 points) j = A[i] - 1 x3000 x3001 x3002 x3003 9. LEA ADD LDR ADD R2 x017 R3, R2, R0 R4, R3, #0 R1, R4, #-1 R2 contains address x3001 + x017 = x3018 (5 points) A[i + j] = A[i] + A[j] x3000 LEA R2 x017 R2 contains address x3001 + x017 = x3018 x3001 x3002 x3003 x3004 x3005 x3006 x3007 ADD LDR ADD LDR ADD ADD STR R3, R4, R3, R5, R5, R3, R5, R2, R3, R2, R3, R4, R3, R3, R0 #0 R1 #0 R5 R0 #0 R4 contains A[i] R5 contains A[j] R5 contains A[i] + A[j] R3 has address (A + j) + i A[i+j] = A[i] + A[j]
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