Question

Question: Consider the following code:

lbu \(t0, 0(\)t1)

sw \(t0, 0(t2)

Assume that the register \)t1 contains the address and the register \(t2 contains the address

. Note the MIPS architecture utilizes big-endian addressing. Assume that the data (in hexadecimal) at the address is . What value is stored at the address pointed to by register \)t2?

Short answer

The value stored at the address pointed to by register $t2 is

Step-by-step solution

Determine ASCII and Binary numbers in MIPS.

The ASCII digits are presented as strings and each digit is 8 bits long. MIPS provides the instructions to load and store bytes. The “Load byte” instruction will load the byte from the memory. The byte will be placed at the 8 extreme right bits of a register. The “Store word” instruction will take the byte from the register and write it to the memory.

Determine the value of $t2.

It is given that the register $t1 contains and$t2 contains .

The MIPS architecture follows a big-endian address, which means the MSB is stored first.

The data at $t1 is

The instruction lbu $t0,0($t1) will load the unsigned bytes to the 8 extreme-right bits of the address $t1.

The instruction sb $t0, 0($t2) will store the value at the specified address.

We know that is the data in $t1, now after the execution of instructions, the consecutive bytes of the word will be stored in four addresses.

Address	data
0x1000 0000	0x11
0x1000 0001	0x22
0x1000 0002	0x33
0x1000 0003	0x44