Remove leading zeroes binary - sparc

Question

I want to basically remove my leading zeroes. When I print out a number for example 17 is 00000 0000 0000 0000 0000 0000 00001 0001 but to do remove those leading zeroes. Because in sparc machine that is what is printed out and I need to do this using some sort of loop or logic or shift function.
this is my psuedocode for printing the binary
store input, %l1 ! store my decimal number in l1
move 1,%l2 !move 1 into l2 register
shift logical left l2,31,l2 !shift my mask 1 31 times to the left
loop:
and l2,l1,l3 ! do and logic between l1 and l2 and put this in l3
compare l3,0 compare l3 zero
bne print 1 !branch not equal to zero, to print 1
if equal to 0
print zero
print 1:
print a 1
go: increment counter
compare counter 32
if counter less than 32 return to loop
shift l2 to the right to continue comparison
so this is what is being done say my input is l1 is 17
00000 0000 0000 0000 0000 0000 00001 0001
10000 0000 0000 0000 0000 0000 00000 0000 and my mask 1 shift left 31 times
this pseucode print out my input decimal into binary. But how can I make it remove leading zeroes?
because in the sparc 17 input inside the machine is
0000 0000 0000 0000 0000 0000 0001 00001

Answer 1

You create the labels, like go and print 1 (more commonly done in all caps and without spaces, FYI). So, starting with bne you should always be printing 1, or falling through to see if it needs to print the 0:
! same initialization
mov 0, l4 ! Initialize a flag to avoid printing
LOOP:
and l2, l1, l3 ! do and logic between l1 and l2 and put this in l3
cmp l3, 0 ! Is this a 0 digit?
bne ALWAYS_PRINT ! If it's not 0, then it must be 1 (was "bne print 1")
cmp l4, 1 ! Should we be printing the 0?
be PRINT_VALUE ! Yes, we need to print the 0 because we have seen a 1
ba INCREMENT ! We should not be printing the 0, so check the next
! digit (ba is "branch always")
ALWAYS_PRINT: !
mov 1, %l4 ! Note that we want to always print for the
! rest of the loop
PRINT_VALUE: ! Do whatever you're doing to print values
print value in l3 ! Always print the value
INCREMENT: ! Formerly your "go:" label
! same logic
! AFTER LOOP IS DONE LOGIC
cmp l4, 0 ! If the flag was never set, then the value is 0
! Alternatively, you could just compare the value to 0
! and skip the loop entirely, only printing 0 (faster)
bne DO_NOT_PRINT ! If it was set (1), then do nothing
print zero ! If it wasn't set, then print the 0
DO_NOT_PRINT:
To walk through it a little, you need to continue to initialize your values and shift the bits to figure out what the current digit is for each iteration. Since you will need another flag, then you need to use another register that is initialized to an expected value (I chose 0, which commonly represents false).
Get current digit into l3 (0 or 1)
See if it is 0
If it's not 0, then it must be 1. So go remember that we found a 1, for later, then print the value and increment/loop.
If it's 0, then see if we have found a 1 before. If so, then print the value and increment/loop. If not, then increment/loop.
For actually printing, I have no idea what you are actually doing. However, you can avoid a second comparison by using the labels. For example, ALWAYS_PRINT will always be used when the value is 1, so you can just set the flag and immediately print 1, then jump to INCREMENT. If you did that, then PRINT_VALUE would only be used to print 0, which could then fall through to INCREMENT.
From a high level language's perspective, you want:
int l2 = // value...
bool seenOneAlready = false;
if (l2 != 0)
{
// MSB first
for (int i = 31; i > -1; --i)
{
int l3 = (l2 >> i) & 1;
if (l3 == 1)
{
seenOneAlready = true;
printf("1");
}
else if (seenOneAlready)
{
printf("0");
}
}
}
else
{
printf("0");
}