So I understand somewhat what the command sll is doing, as I read this and its pretty much just shifting all the bits left by 1. I'm just wondering why would I do this?
I have an assignment from class with an example of it... Where $s6 and $s7 is the base address for an array and $s1/$s2 are just some variables.
sll $t0, $s0, 2
add $t0, $s6, $t0
sll $t1, $s1, 2
add $t1, $s7, $t1
...
Why shift a bit? What is it doing in simple terms? My first thought it has something to do with indexing the variables in the array...but I'm not sure.
its pretty much just shifting all the bits left by 1
The example they showed was a shift by 1 bit. The sll instruction isn't limited to just shifting by 1 bit; you can specify a shift amount in the range 0..31 (a shift by 0 might seem useless, but SLL $zero, $zero, 0 is used to encode a NOP on MIPS).
A logical left shift by N bits can be used as a fast means of multiplying by 2^N (2 to the power of N). So the instruction sll $t0, $s0, 2 is multiplying $s0 by 4 (2^2) and writing back to $t0. This is useful e.g. when scaling offsets to be used when accessing a word array.
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