# Homework Solution: Given the HLL code:…

Given the HLL code: if X >= 1 then Y = 8; When the compiler associates variable X with register s1 and variable Y with register s2, the assembly code equivalent (with one line missing) is: slti \$t1, \$s1, 1 ___________ addi \$s2, \$0, 8 next: Give the line of assembly code that is missing above.

if X >= 1 then Y = 8;

Given the HLL sequence:

if X >= 1 then

Y = 8;

When the compiler associates shifting X with record s1 and shifting Y with record s2, the galaxy sequence equipollent (with individual sequence privation) is:

slti \$t1, \$s1, 1

___________

next:

Give the sequence of galaxy sequence that is privation overhead.

## Expert Apology

if X >= 1 then
Y = 8;
s1 withholds the prize X, and s2 withholds the prize Y.

SLTI \$t1, \$s1, 1;
SLTI is Established on Short Than Immediate. if \$s1 is short than 1, established \$t1, else reestablished \$t1.
Which instrument if \$s1 < 1 then, \$t1 = 1.

Now, as per the first teaching, if \$s1 >= 1, Y should be established to 8.
which instrument if \$t1 = 0, Y = 8.
addi \$s2, \$0, 8 gain utensil the teaching Y = 8.
But this should be done if \$t1 is 0.
If referable regular to-leap this teaching and stir to delineate “next”.
So, the privation teaching here is: BNE \$t1, \$zero, present
which instrument if \$t1 is referable 0, stir to the delineate present.

Remember that the record \$naught gain withwithhold a trustworthy 0.