Q: Assistance in understanding and solving these fill in the blanks example on D
ID: 3746008 • Letter: Q
Question
Q: Assistance in understanding and solving these fill in the blanks example on Digital Design and Computer Architecture with the steps of the solution to better understand, thanks.
*Using MIPS The MIPS language. The book Digital Design and Computer Architecture covers it.
a) The byte address of memory word 1510 is 0x_________ in a byte-addressable memory.
If the number 0xFF223344 is stored at word 1510 in a big-endian machine then the byte addressed by 0x3D has the data value 0x_______________ .
b) Convert the three instructions below into their hexadecimal sequence (i.e. hexadecimal version of their binary sequence representations in memory)
1. add $t0, $s0, $s1 in hexadecimal is 0x_________ .
2. lw $t0, 0x20($t7) in hexadecimal is 0x__________.
3. addi $s0, $0, 10 in hexadecimal is 0x_________.
c) Suppose that $s3 initially contains 0x23456789.
The instruction sw $s3, 4($0) stores the data in register $s3 into memory word number ____________ . This word has the byte address 0x_________ .
After this sw instruction has stored in memory, the instruction lb $s3, 4($0) loads data into $s3. In a big-endian machine the value inside $s3 will be 0x and in a little-endian machine the value will be 0x____________ .
Explanation / Answer
ANSWER:
b)
Convert the three instructions below into their hexadecimal sequence (i.e. hexadecimal version of their binary sequence representations in memory)
1. add $t0, $s0, $s1 in hexadecimal is 0x02114020
Explanation:-
R-type instruction format:-
Opcode(6-bit) Rs(5-bit) Rt(5-bit) Rd(5-bit) shamt(5-bit) func(6bit)
000000 $s0(10000) $s1(10001) $t0(01000) 00000 100000
Convert into hex->0X02114020
2. lw $t0, 0x20($t7) in hexadecimal is 0x8de80020
Explanation:-
I-type instruction format:-
Opcode(6-bit) Rs(5-bit) Rt(5-bit) address(16-bit)
100011 $t7(01111) $t0(01000) 0000000000100000
Convert into hex->0X8de80020
3. addi $s0, $0, 10 in hexadecimal is 0x2010fff6
Explanation:-
I-type instruction format:-
Opcode(6-bit) Rs(5-bit) Rt(5-bit) Immediate(-ve 2's complement10)
001000 $0(00000) $s0(10000) 1111111111110101
Convert into hex->0X2010fff6
(C)
Explanation:-
I-type instruction format:-
Opcode(6-bit) Rs(5-bit) Rt(5-bit) address(16-bit)
101011 $t0(00000) $s3(10011) 000000000000000100
Convert into hex->0Xac130004
---------------------------------------------------------------
I-type instruction format:-
Opcode(6-bit) Rs(5-bit) Rt(5-bit) address(16-bit)
100100 $t0(00000) $s3(10011) 000000000000000100
Convert into hex->0X80130004
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