finalexam_2002 solutions

finalexam_2002 solutions - C Opcode : ADD, AND (one is...

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Department of Electrical and Computer Engineering University of Texas at Austin EE 379K Fall 2000 Y. N. Patt, Instructor TAs: Kathy Buchheit, Laura Funderburg, Chandresh Jain, Onur Mutlu, Danny Nold, Kameswar Subramanian, Francis Tseng, Brian Ward Solutions to Exam 2 Problem 1 Part 1 Both instructions are noops (they do *essentially* nothing) Difference: B sets the condition codes. Part 2 128 locations on a page, 8 pages Part 3 LDI R6, x50 Part 4 largest address for current LDR : x323F largest address for modified LDR : x321F smallest address for modified LDR : x31E0 Problem 2 A Opcode : TRAP Function : Carries the trapvector, which is used to find the starting address of the system call. B Opcode : LDR, STR, JSRR (one is enough) Function : Carries the contents of the base register.
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Unformatted text preview: C Opcode : ADD, AND (one is enough) Function : Selects whether an immediate operand or a value from a register will be used (Bit 5). Problem 3 Part A P condition code. Part B Whether the instruction sets the condition codes. Problem 4 1100 1 00 001 000100 Problem 5 1111 0000 00100111 No. Because GETC and OUT overwrite R7 linkage to main program will be lost. Problem 6 Outputs the input characters in reverse order. Problem 7 Filled blanks in order: ADD R1, R1, #-1 LDR R4, R1, #0 ADD R0, R0, #1 ADD R1, R1, #-1 BRnzp LOOP Problem 8 Part A Sets the Interrupt Enable bit of KBSR and continually outputs 2. Part B Echoes the input character and halts. Part C A number of 2's followed by the key struck by the user....
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finalexam_2002 solutions - C Opcode : ADD, AND (one is...

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