sll 的目的是什么，并在翻译后的 MIPS 代码中添加？

Question

I was reading some MIPS notes regarding translating C statement to MIPS.

C code of swap function below:

swap ( int v[], int k) {
    int temp;
    temp = v[k];
    v[k] = v[k + 1];
    v[k + 1] = temp;
}

I was told that k maps to $5, base address of v[ ] is mapped to $4 and temp is mapped to $15. The swap function takes in the arguments k = 3 and assume that base address of v is 2000.

Simplified MIPS version below:

swap:

    sll $2, $5, 2
    add $2, $4, $2
    lw  $15, 0($2)
    lw  $16, 4($2)
    sw  $16, 0($2)
    sw  $15, 4($2)

The confusion here is why is there sll and add in the translated MIPS?

Answer 1

This comes from being byte addressable , as most modern machines are.

Byte addressing means that the processor can access each individual byte of memory — in such a system, incrementing a valid byte address value by 1 means referring to the next sequential byte of memory.

Words, used by the int datatype, are 32-bits wide, so require 4 bytes of memory. Thus, a[0] occupies bytes addr+0 , addr+1 , addr+2 , and addr+3 , where addr is a byte address of a . a[1] is offset from a[0] by 4 bytes, at addr+4 !

C knows that the elements of "arrays of int " require 4 bytes each so 4 bytes separation, and thus it knows that a+i under the hood means a + i * 4 , which will refer to 4 consecutive addresses for each element of the array, starting at i*4 , which leaves room for all the lower elements.

The computation i * 4 is sometimes called "scaling". Whereas in C we can refer to the index position directly a[i] , in assembly we have to scale the index explicitly.

The sll is an efficient way to multiply by 4 — the constant 2 in the sll means 2 digits to the left (aka: * 100 ₂ , ie * 100 in binary).

The add sums the array base with the scaled index to make the actual address of the desired word for the array reference.

We can also see the generated code using constants 0 or 4. Because these constants are also scaled, the 4 refers to the next element of the array beyond what was computed. Which is to say that if 0($2) refers to a[k] then 4($2) refers to a[k+1] .