Loading and storing bytes in MIPS

Question

I'm studying for an exam tomorrow and I'm quit confused on the loading/storing bytes topic. I have this example: I don't understand how he got the answers in red at all. Could someone help explain this to me?

Accepted Answer

```
add    $s3, $zero, $zero
```

This performs the addition `$s3 = 0 + 0`, effectively setting the register `$s3` to a value of zero.

```
lb     $t0, 1($s3)
```

This loads a byte from a location in memory into the register `$t0`. The memory address is given by `1($s3)`, which means the address `$s3+1`. This would be the 0+1=1st byte in memory. Since we have a big-endian architecture, we read bytes the 4-byte chunks "big end first".

```
byte:  0   1   2   3
      00  90  12  A0
```

The 0th byte is `00`, and the 1st byte is `90`. So we load the byte `90` into `$t0`.

```
sb     $t0, 6($s3)
```

This stores a byte from the register `$t0` into a memory address given by `6($s3)`. Again this means the address `$s3+6`.

```
byte:  4   5   6   7
      FF  FF  FF  FF
```

becomes

```
byte:  4   5   6   7
      FF  FF  90  FF
```

Now, what if the architecture was little-endian? This would mean bytes are arranged "little end first" in memory, so the effect of the 2nd and 3rd instructions change.

```
lb     $t0, 1($s3)
```

This loads the byte in memory address 1 into register `$t0`. But now the addresses are "little end first", so we read `12` into the register instead.

```
byte:  3   2   1   0
      00  90  12  A0
```

Next...

```
sb     $t0, 6($s3)
```

This stores the byte in register `$t0`, which is `12` into a memory address 6. Again with little-endian architecture:

```
byte:  7   6   5   4
      FF  FF  FF  FF
```

becomes

```
byte:  7   6   5   4
      FF  12  FF  FF
```

Loading and storing bytes in MIPS

Solution

Problem