Load 32-bit constant to register in MIPS

Question

I was confused about this part while I study MIPS.

The textbook written by Professor John L. Hennessy say if we get some big constant to load, we should

lui $s0, upper(big)
ori $s0, $s0, lower(big)

But why don't we just do

addi $s0, $zero, big

Since the registers are 32-bit, this is more strightforward, isn't it?

score 10 · Accepted Answer · edited Aug 14 '19 at 16:43

10

The immediate argument passed to addi is only 16 bits. To load a 32-bit immediate value that is outside the range of a 16-bit value you need to do it in two goes, as in the example from your text book.

(Anticipating a further question, the reason there is no load immediate or add immediate instruction which takes a 32-bit immediate value is because the MIPS ISA uses fixed size 32-bit instructions, so there are always < 32 bits available for any instruction arguments - this is very common in RISC architectures.)

edited Aug 14 '19 at 16:43

phuclv

37,963
15
156
475

answered Oct 31 '12 at 15:06

Paul R

208,748
37
389
560

1

Thank you! Because of the fixed length, the immediate format leaves only 16 bit for constant, other 16 bits for the operation and register code. – lucasKo Nov 01 '12 at 13:58
Another question regarding combining "lui", "ori" and "or" to load a large constant into a register. 1. Why would you do this rather than just computing the value with add-operations and save it into a register? 2. Also, how do you use a constant bigger than 26 bits? A j-type instruction takes only a 26 bit constant. – Energizer1 Dec 22 '19 at 17:32

score 6 · Answer 2 · answered Oct 31 '12 at 15:07

6

Yes registers are 32-bits but how can you specify a 32-bit number in an instruction that is 32 bits? An instruction consists of opcode and data. So, you can't squeeze an opcode + 32-bit data in a single addi.

answered Oct 31 '12 at 15:07

yaman

759
3
17

score 1 · Answer 3 · answered Mar 29 '18 at 02:39

1

Alternatively, using .data and .text

.data
   word32bits: .word 0xFFFFFFFF

.text
   lw $t0, word32bits # $t0 now contains your 32 bit word.

answered Mar 29 '18 at 02:39

Rafael Nagel

23
1
5

You can do this, but lw is a relatively expense operation, so it's better not to. – Ellen Spertus Aug 11 '20 at 00:44

score 0 · Answer 4 · edited May 23 '17 at 12:02

0

You can use literal pool if you need to load a lot of constants. Then each constant load costs only 1 load instruction. More information here and here

edited May 23 '17 at 12:02

Community

1
1

answered Aug 01 '13 at 13:52

phuclv

37,963
15
156
475

Load 32-bit constant to register in MIPS

4 Answers4

Linked