Simple MIPS question, about load byte

Question

I have the following question here

.data
a: .asciiz "2021"
x: .byte 7,2,12
.text
main: addi $t2, $0, 1
lb $t3, a($t2)

Can someone explain to me, HOW the value of $t3 is 48?

thanks EDIT this is an another question which is similiar, and is confusing.

.data
a: .word 12,-5,4,0
x: .byte 5
.text
main: addi $t1, $0, 8
lw $t2, a($0)
lw $t3, a($t1)

How will you load word, from index 8, when 'a' has a length of 4?

Answer 1

Yes, when you add $0 and 1, you get 1, which is put into $t2.

Then, when you evaluate a($t2), that's the second byte (offset 1 since it's based at offset 0) of a which is the "0", ASCII code 0x30 or 48.

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From various pieces of information:

ADDI -- Add immediate (with overflow)
Description:
    Adds a register and a sign-extended immediate value
    and stores the result in a register
Operation:
    $t = $s + imm; advance_pc (4);
Syntax:
    addi $t, $s, imm

LB -- Load byte
Description:
    A byte is loaded into a register from the specified address.
Operation:
    $t = MEM[$s + offset]; advance_pc (4);
Syntax:
    lb $t, offset($s)

Register $0 always contains the hardwired value 0. MIPS has established a set of conventions as to how registers should be used. These suggestions are guidelines, which are not enforced by the hardware. However a program that violates them will not work properly with other software.

Those little snippets should hopefully be enough to explain what it's doing.

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And, regarding your edit, you're incorrectly thinking that .word 12,-5,4,0 has a length of 4 bytes. In fact it has a length of 16 bytes since words in MIPS are 32 bits (four bytes) wide.

So when you load from byte offset 8, you will get the word 4.