Question

1. It is desired to load A[10] into $8, i.e., "lw $8, X($9)". Assume $9 contains the base address of array A. Address Memory content lw $8, X($9) 0x0001 0000 Ox0001 0004 0x0001 0008 Ox0001 000c 0x0001 00101 Ox0001 0014 0x0001 0018 Ox0001 001c 0x0001 00201 0x0001 0024 0x0001 0028 A[O] A[1] $9 contains the based address of array A (i.e., $9 = 0x0001 001c) (a) What is the memory address for A[10]? Answer in hexadecimal number. Hint: Remember that A[0] is at 0x0001 001c, A[1] is at Ox0001 0020, etc. (b) What should be the offset (X) in the "lw" instruction? Answer in hexadecimal number. Hint: The location of A[10] is equal to $9+X. What is X? (c) What is the machine code representation of the "lw $8, X($9)" instruction? Opcode for lw = 10 0011. Answer in binary and hexadecimal format. The format of the "lw" instruction is as follows. 0x23 rs r t 16-bit offset

Answer 1

solution:

Solution: @ Addoers of A[O] - 0x0001 0010 A[]0x0001 0020 A[2] = 0 x0001 0024 ^ [3] = 0X0001 0028 Alu] : 0x0001 0020 A[s] = 0X0001 0030 A C6] 0x0001 0034 A[77 - 0x0001 0038 A[c] = 0X000] 003c A[9] = 0x000) 0040 A [10] = 0X00010044 Address to A[10] = 0x 000] 0044 ☺ the location of A{o] = $9+X location of $9 = 0x0001 0010 x=0x00010044 -0X00010010 1x: 0 X 0000 00.25

© w $8,x(89) in Mips format opcode is rt 16 bit offset 100011 01001 1000 0000 0000 0010 0101 Here source register =$9 [write qin binary folo sbits tempolary registers = $8 (8'in binary for sbits] L6 bit otiset - XS0025 Ti-y0101 0010 J 00000000 0010

please give me like..its help me to write more questions..thank you..