Question

Assembly Machine Programming/Coding

Why are the values in $7 different from $6? addi $8, $0, -3 addi $9, $0, 8 slt $7, $8, $9 sltu $6, $8, $9 mult $8,99 What's the difference between mult vs. multu instructions? Consider ($11, $10) vs. ($13, $12). mflo $10 mfhi $11 multu $8, $ 9 mflo $12 mfhi $13

Answer 1

A. Slt is nothing but set if less than instruction(signed).

slt $d, $s, $t

If ($s < $t) $d =1 else $d =0

Whereas sltu also does the same operation but It treats the values as unsigned.

In the given example $8= -3( 11111101)and $9=8(00001000)

as -3<8 , $7 =1

In sltu, comparison will be

11111101(253 in decimal) is not less than 8, hence $6 is not set.

This is the difference between the two instructions slt and sltu

B. Mult  instruction operates on signed numbers and multu operates on unsigned numbers.

During multiplication whether it is signed or unsigned, the lower 32 bits of src1 * src2 is stored in lo register and upper 32 bits of src1 * src2 is stored in hi register.

The difference between signed and unsigned multiplication is overflow, the lo(lower 32 bits of result)  part will remain same. But the hi part(upper 32 bits of result) will vary in such a way, with signed multiplication, the negative result is sign extended. But in unsigned multiplication negative result is not sign extended.

Thus in the above instructions,

mult $8, $9 is a signed multiplication, the lower 32 bits of this multiplication result is copied to register $10 and upper 32 bits are copied to register $11.

multu $8, $9 is an unsigned multiplication, the lower 32 bits of this multiplication result is copied to register $12 and upper 32 bits are copied to register $13.

As stated above the the lower 32 bits of the results will be same for both signed and unsigned multiplication. Thus registers $10 and $12 will hold the same values after execution of the instructions. Whereas the registers $r11 and $r13 will have different values based on sign extension on negative value of result as stated above.