Q: Am asking for assistance in understanding and solving these fill in the blank
ID: 3746081 • Letter: Q
Question
Q: Am asking for 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.
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
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
2. lw $t0, 0x20($t7) in hexadecimal is 0x8DE80020
3. addi $s0, $0, 10 in hexadecimal is 0x2230FFF6
c) Suppose that $s3 initially contains 0x23456789.
The instruction sw $s3, 4($0) stores the data in register $s3 into memory word number 0x0004 . This word has the byte address 0x0010 .
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 0x89 and in a little-endian machine the value will be 0x23 .
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