I am creating a program for a user to enter an integer and then check each bit and count how many 1's is in it's binary value. So if I input 4673 I should get "4" as an output because there is 4 ones. This is the code I have below, for some reason I am only getting "0" as an output. My guess is I am not properly loading each bit with the "andi" and "srl". I check it step by step and when it comes to andi and srl $t0 never holds a value of 1, so I must not be shifting bit by bit?

 .data
Msg: .asciiz "Enter an integer: "



.text
 # Print the first message
 li $v0, 4
 la $a0, Msg
 syscall

 # Prompt the user to enter the first integer
 li $v0, 5
 syscall

 # Store the first integer in $t0
 move $t0, $v0

 addi $t3, $zero, 1

main:
 bgt $t3, 31, exit
 addi $t3, $t3, 1
 andi $t0, $v0, 1
 srl $t0, $t0, 1
 bne $t0, $zero, count 
 j main


count:
 addi, $t1, $t1, 1
 # Shift to the next bit and then go back to main
 j main
 
exit:

# Tell the interpreter to get read to print an integer
 li $v0, 1
 add $a0, $zero, $t1
 
 #Print the integer
 syscall
 
 # End the program
 li $v0, 10
 syscall

CodePudding user response：

You've got this instruction andi $t0, $v0, 1 in your loop. But $v0 never changes within the loop, so you're always getting the same value. And regardless of whether that values was 0 or 1, it's going to be 0 after the srl on the following line.

The whole bit-counting loop could be replaced by something like this:

li $t1, 0  # number of 1-bits
count_ones:
    andi $t2, $t0, 1            # t2 = input & 1
    addu $t1, $t1, $t2          # count  = (input & 1)
    srl $t0, $t0, 1             # input >>= 1
    bne $t0, $zero, count_ones  # loop until no 1-bits left

Note that there are more efficient ways of doing this, without any loops at all. See How to count the number of set bits in a 32-bit integer?