CODEWITHSUNDEEP
assemblymips
shawn a
shawn a

Reputation: 809

MIPS counting digits in unsigned integer

A user enters a binary number(ie: 110100), I need to calculate the number of consecutive 0's starting from the LSB, so it would be 2 in the case of 110100.

My problem is counting the length of this entered string. But I found this to be impossible because im checking by way of

      beqz $t2,finished

which will immeadiately stop at the first 0 at 110, even though we know this is not the end of the string.

My attempt:

    ## Read string from user ##
    la $a0, str
    li $a1, 100
    li $v0, 8
    syscall
    ##

    ## Loop each byte in string until null-terminator is found ##
    loop:
        move $t0,$t1
        add $t0,$t0,$a0 # current address($t0)=counter*1(size of character in bytes)+RAM address of start of array

        lb $t2,($t0)    # load next byte in string 

        lw $s0, del

        beqz $t2,finished   # if current character==null-terminator, exit loop

        addi $t1,$t1,1  # increase counter

        beqz $t2,addtoh

        j loop  # loop around

    addtoh:
        addi $t3,$t3,1 #
        j loop  # loop around
    finished:
    ##

#################################################
#                                               #
#               data segment                    #
#                                               #
#################################################

            .data
        str:    .space 100
        endl:   .asciiz "\n"

Upvotes: 1

Views: 3203

Answers (1)

RobertB
RobertB

Reputation: 1076

I don't see where you're initializing $t1. Unlike high-level languages, machine code doesn't initialize memory or registers automatically. $t1 could be starting out anywhere. So when you do this:

move $t0,$t1    # unknown 32-bit value loaded into $t0
add $t0,$t0,$a0 # $t0 is now an unknown number of bytes past (or before!) the start of the buffer
lb $t2,($t0)    # load next (random) byte in string

... you could end up looking at bytes anywhere in the memory space. And since a lot of those bytes will end up being zero, your code will finish sooner than you expected.

Upvotes: 1

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