CODEWITHSUNDEEP
mipsbit-shift
eChung00
eChung00

Reputation: 633

How does the sll instruction in MIPS work?

The following assembly code is given in my text book.

Loop:
sll $t1, $t0, 2 
add $t2, $a0, $t1
sw $zero, 0($t2)

addi $t0, $t0, 1
slt $t3, $t0, $a1
bne $t3, $zero, Loop

# return where we were
jr $ra

From this code I have two question to ask.

The first question is about the second line from the top. I get that the instruction sll is shift left logical which shifts bit to the left. Since the shift amount is 2, it will make 0000 -> 0100 = 4 in decimal. But I don't get it after the first loop. If we shift this to the left by 2, isn't it multiplied by more than 4??

And the second question is if it is possible to optimize this code?? In my opinion, I can modify sll and add parts in the code but I am not sure. Any comment??

Upvotes: 0

Views: 9231

Answers (3)

Wilson
Wilson

Reputation: 251

Consider the following binary:

0000 0001

If you shift the bits left by 1 digit you get:

0000 0010

If you shift again to the left by 1 digit:

0000 0100

And again:

0000 1000

The binary values above are equivalent to 1; 1x2=2; 2x2=4; 4x2=8. Shifting bits to the left is multiplying the value by 2^N if N is the number of bits you are shifting.

Another example of shifting:

Assume $t1 contains 0000 1111

sll $t0, $t1, 3    # $t0 = $t1 * 2^3

Now $t0 contains 0111 1000

You can verify this by performing decimal multiplication. 15 * 8 = 120.

Upvotes: 0

Michael
Michael

Reputation: 58437

[is it] possible to optimize this code?

A more compact way of doing the same thing would be:

sll $a1, $a1, 2
addu $a1, $a1, $a0   # $a1 = $a1 * 4 + $a0
Loop:
sw $zero, ($a0)
addiu $a0, $a0, 4
bne $a0, $a1, Loop

I'm making these assumptions:

  • The original values of $a0 and $a1 are not needed anymore after the loop ends. If they are needed, save the original values somewhere (in other registers or on the stack) before entering the loop, and restore them afterwards.
  • $t0 starts out at zero. If not, you'll have to add $t0 * 4 to $a0 before the loop. I'm also assuming that the value of $t0 after exiting the loop is irrelevant.

Upvotes: 0

phuclv
phuclv

Reputation: 41774

Shifting left will insert 0's, not 1's. So 0000 would still be 0000, 0001 will become 0100 after the shift

Upvotes: 1

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