MIPS Computer Architecture Question

Requirments to Answer: Solid Foundation of MIPS Programming / Hexidecimal formatting / Byte Ordering

Asking: Good Explanation and Solution

1. (3 pts) Assume that the instruction j NEXT is at address ox0ODAEO5C, and the label NEXT is at address OxOODAFA28. Then, the 26-bit immediate stored in the jump instruction for the labe NEXT is 2. (3 pts) Assume that the instruction beq $to, $tl, NEXT is at address 0x04DAE05C, and the label NEXT sat address 0x04DAFA28Then, the 16-bit immediate stored in th branch instruction is 3. Consider the following data definitions data varl: .byte Z, 1, 2. 5. B var2: .half -5. xDfC var3: word 0x12345678, Oxff align 3 stri: .asciiz "My Stringin a) Show the content of each byte of the allocated memory, in hexadecimal for the above data definitions. The Little Endian byte ordering is used to order the bytes within words and half words. The ASCIl code of character A is 0x41, and 'o is 0x30. Indicate which bytes are skipped or unused in the data segment. Data Segment Symbol Table Address ByteByte Byte Byte 0x10010000 0x10010004 0x10010008 0x1001000C 0x10010010 0x10010014 0x10010018 0x10010010 0x10010020 0x10010024 0x10010028 0x10010020 Label Address b) Construct a symbol table showing the symbols and their corresponding addresses in hexadecimal. c) How many bytes are allocated in the data segment including the skipped bytes