Unfaking Direct-Mode Loads and Stores

Some Interesting Variants of the Load Instruction

Instruction Load immediate (fake) Load upper immediate Load address (fake)

Example in assembly language li $10, 17 lui $11, 71 la $12, 0x10010044

What it does Place the value 17 into $10 Place the value 71 into the upper half of $11 Place the address 0x10013456 into $12

Assembly language after unfaking ori $10, $0, 17

lui $12, 0x1001

ori $12, $12, 3456

Previous row assembled into I-format instructions

13 0 10 17

15 0 11 71

15 0 12 1001

13 12 12 3456

Result in register

0 17

71 0

1001 3456

The immediate values for li and lui are only 16 bits long but are being placed into a 32-bit register. This requires filling the other 16 bits with zeroes. (The complete truth about this will be evident once we have worked out how to represent negative numbers. In the meantime just remember that there's more to come on that subject.)
We have a 32-bit value in the la instruction that is split into two 16-bit halves. The top half goes into the lui, and the bottom half goes into the ori.
In the unfaking of the la instruction, after the lui executes and before the ori executes register $12 will look like this

1001 0
There are hexadecimal values given in some of the assembly-language code examples. In a real case these values would appear as decimal numerals. The values are given in hex here so that they will obviously match the values in the boxes.
The direct-mode addresses in the la example above and the lw example below are given in hex for the same reason. Ordinarily they would be given as labels from the .data segment of the program and the values would be looked up in the symbol table. (e.g. lw $13, x)

Now we need to talk about unfaking the direct-mode load instruction. Suppose the instruction is

lw $13, 0x10015678

The unfaking is a sequence of two instructions

lui $1, 0x1001

lw $13, 0x5678($1)

What this sequence will do is to make register $1 look like this

1001 0

and then use it as the base register to calculate the effective address for the load instruction. The effective address is the sum of $1 and 0x5678. Since 0x5678 is a 16-bit value, it fits into the bottom half of the sum, yielding

1001 5678

which is, of course 0x10015678, the value given in the instruction.

Again, the value 0x10015678 is split into two 16-bit halves, and each half appears in one of the two instructions that unfake the fake.

15 0 1 1001

35 1 13 5678