Moving values around

This program keeps a rectangle's width and height in registers. It calculates the perimeter with 2 × (width + height), makes two copies of the result, and finds the difference between the two sides. Nothing is read from memory.

.text
main:
    li t0, 30           # width = 30
    li t1, 12           # height = 12
    add t2, t0, t1      # width + height = 42
    add t2, t2, t2      # perimeter = 2 * 42 = 84

    mv t3, t2           # copy the perimeter
    add t4, t2, zero    # copy it again
    sub t5, t0, t1      # width - height = 18

In li t0, 30, t0 is a register name and 30 is the number placed in that register. After the first two lines, t0 holds 30 and t1 holds 12.

An add names its destination first, followed by its two sources. add t2, t0, t1 reads t0 and t1, then writes their sum to t2. It does not change either source register. The next add uses t2 as both sources, so it adds 42 to itself and writes 84 back to t2.

mv t3, t2 copies the value in t2 to t3. The following line makes another copy in t4: zero always supplies the value 0, so adding it does not change the value. mv is a pseudo-instruction that the assembler expands to addi t3, t2, 0; it has the same copying effect.

The final sub also names its destination first. It calculates t0 - t1, or 30 - 12, and puts 18 in t5. After the last instruction, the registers hold:

register	value	meaning
t0	30	width
t1	12	height
t2	84	perimeter
t3	84	copied perimeter
t4	84	copied perimeter
t5	18	width minus height

Try it

Change the width to 7 and the height to 5. Before running the program, predict the final values of t2, t3, t4, and t5. Then run it and compare the registers with your prediction.

Show answer

You should get t2 = 24, because 2 × (7 + 5) = 24. Both copies have the same value, so t3 = 24 and t4 = 24. The subtraction gives t5 = 2.