Loop Body

The address of the byte in the lbu instruction is computed by (displacement + $9). Since the base, $9, has the address of (points at) the character, the displacement is zero. The lbu loads the character into the low-order byte of $10. The upper three bytes are zero. The beq instruction tests if the entire register is zero, but, of course, that also tests if the low-order byte is zero.

 


## strlen.asm
##
## Count the characters in a string
##
## Registers:
##  $8 -- count
##  $9 -- pointer to the char
## $10 -- the char (in low order byte)

        .text
        .globl  main
		
# Initialize
main:    ori      $8,$0,0       #  count = 0
         lui      $9,0x1000     #  point at first char

# while not ch==null do
loop:    lbu   $10,0($9)        # get the char
         sll   $0,$0,0          # load delay
           
         beq   $10,$0,done      # exit loop if char == null
         sll   $0,$0,0          # branch delay

         addiu    ,,   # count++

         addiu    ,,   # point at the next char

         j       loop
         sll   $0,$0,0          # branch delay slot
         
# finish
done:    sll   $0,$0,0           # target for branch

         .data
string:  .asciiz  "Time is the ghost of space."

Next, the program must increment the count, then move the base register (the character pointer) to the next byte of the string.

Answer:

Loop Body

QUESTION 5: