confusion in MIPS storing bits, - binary

Question

hi im trying to store a bit into a temp. register. Am i doing this correctly? and while i am at it, i am trying to see how many 1's are in the binary forms of a decimal number (0-16) am i doing this right? here is the chunk of code that matters, the rest works fine(just output and what not)
# for (i = 0; i <= 16; i++)
li $s0, 0 # i = 0
li $s3, 0 #constant zero
li $s4, 0 #i-2=0
bgt $s0, 16, bottom
top:
# calculate n from i
# Your part starts here
sb $t1, ($s0) #store LSB from number i in t1
sltu $t7,$s3,$t1 # check if t1 is a 1, if it is set t7 = 1
add $s1 ,$s3,$t1 # add 1 to n
ble $s4, 7, bloop # loop through rest of bits
bloop:
srl $t7, $s0, 1 # move bits in i right one, bringing in a zero from the left
sltu $t6, $s3, $t7 # check if t7 is a 1, if it is set t6 to 1
add $s1, $t6, $s1 # add increment n up 1
bgt $s4, 7, continue # break out of loop
continue:

Answer 1

Lets look at your code:
li $s0, 0
li $s3, 0
li $s4, 0
these instructions set the 3 registers $s0, $s3, and $s4 to 0
bgt $s0, 16, bottom
Branch to bottom if $s0 is greater than 16. Now $s0 will be 0 initially, and you have not labels before this, so there's no way to get back here, so this branch will never be taken. Also, it won't even assemble, as you never define the bottom label anywhere.
top:
The label top, so you can branch back to here as a loop -- but your code has no branches to top, so it never actually will.
sb $t1, ($s0)
store a byte from $t1 at the address given by $s0. Now the first time into the loop, $t0 has never been set to anything, so what are you storing? Worse, $s0 has been set to 0, so you're storing at address 0, which will crash (give a runtime error).
So this store really makes no sense. It's storing garbage to an invalid address.
sltu $t7,$s3,$t1
This compares $s3 to $t1 and sets $t7 to 0 or 1 depending on whether $s3 is less than $t1. $t1 is still garbage (you've never put anything into it), but it turns out that doesn't matter as $s3 is 0, so the comparison will always be false. Not that it matters anyways, as you never use $t7 for anything after this.
add $s1 ,$s3,$t1
Add $s3 (which is 0) to $t1 (which is garbage), and store the result(garbage) in $s1
ble $s4, 7, bloop
bloop:
Compare $s4 to 7 and branch if it is less than 7. Since $s4 is 0, it will branch, but since bloop is immediately after this, the branch has no effect. So this (and the label) might was well be deleted, as they don't do anything.
srl $t7, $s0, 1
shift $s0 left 1 bit position and store it into $t7. $s0 was set to 0 above, and shifting 0 gives 0, so this stores 0 in $t7.
sltu $t6, $s3, $t7
compare $s3(0) to $t7(0) and store the less-than result in $t6 (0)
add $s1, $t6, $s1
add $t6 to $s1, putting the result in $s1 -- this is the first instruction since the first three li instructions that makes any kind of sense. But as $t6 was zer0, it doesn't actually do anything
bgt $s4, 7, continue
continue:
Another branch that does nothing as the target is right after the branch.
So overally, you code looks like a bunch of randomly chosen instructions that don't do anything sensible. There's no loop, despite comments about looping. There nothing related to "bits" at all, despite title of the question.