Yes.

The details are slightly complicated. But rest assured that there are international standards for this.

The MIPS instruction that loads a word into a register is the `lw` instruction. The store word instruction is `sw`. Each must specify a register and a memory address. A MIPS instruction is 32 bits (always). A MIPS memory address is 32 bits (always). How can a load or store instruction specify an address that is the same size as itself?

An instruction that refers to memory uses a base register and an offset. The base register is a general purpose register that contains a 32-bit address. The offset is a 16-bit signed integer contained in the instruction. The sum of the address in the base register with the (sign-extended) offset forms the memory address.

Here is the load word instruction in assembly language:

```lw   d,off(b)       # \$d <-- Word from memory address b+off
# b is a register. off is 16-bit two's complement.
# (The data from memory is available in \$d after
#  a one machine cycle delay.)
```

At execution time two things happen: (1) an address is calculated by adding the base register `b` with the offset `off`, and (2) data is fetched from memory at that address.

Because it takes time to copy data from memory, it takes two machine cycles before the data is available in register `\$d`. In terms of assembly language this means the instruction immediately a `lw` should not use `\$d`.

### QUESTION 6:

Write the instruction that loads the word at address 0x00400060 into register \$8. Assume that register \$10 contains 0x00400000.

`lw \$8, (   )`