The answer is below.
## 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."
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.
Next, the program must increment the count, then move the base register (the character pointer) to the next byte of the string.
Fill in the blanks.