﻿ Lesson 9 Screen04

# Lesson 9 Screen04

The Screen04 lesson builds on Screen03, by teaching how to manipulate text. It is assumed you have the code for the Lesson 8: Screen03 operating system as a basis.

## 1 String Manipulation

Variadic functions look much less intuitive in assembly code. Nevertheless, they are useful and powerful concepts.

Being able to draw text is lovely, but unfortunately at the moment you can only draw strings which are already prepared. This is fine for displaying something like the command line, but ideally we would like to be able to display and text we so desire. As per usual, if we put the effort in and make an excellent function that does all the string manipulation we could ever want, we get much easier code later on in return. Once such complicated function in C programming is sprintf. This function generates a string based on a description given as another string and additional arguments. What is interesting about this function is that it is variadic. This means that it takes a variable number of parameters. The number of parameters depends on the exact format string, and so cannot be determined in advance.

The full function has many options, and I list a few here. I've highlighted the ones which we will implement in this tutorial, though you can try to implement more.

The function works by reading the format string, and then interpreting it using the table below. Once an argument is used, it is not considered again. The return value of the function is the number of characters written. If the method fails, a negative number is returned.

Table 1.1 sprintf formatting rules
Sequence Meaning
Any character except % Copies the character to the output.
%% Writes a % character to the output.
%c Writes the next argument as a character.
%d or %i Writes the next argument as a base 10 signed integer.
%e Writes the next argument in scientific notation using eN to mean ×10N.
%E Writes the next argument in scientific notation using EN to mean ×10N.
%f Writes the next argument as a decimal IEEE 754 floating point number.
%g Same as the shorter of %e and %f.
%G Same as the shorter of %E and %f.
%o Writes the next argument as a base 8 unsigned integer.
%s Writes the next argument as if it were a pointer to a null terminated string.
%u Writes the next argument as a base 10 unsigned integer.
%x Writes the next argument as a base 16 unsigned integer, with lowercase a,b,c,d,e and f.
%X Writes the next argument as a base 16 unsigned integer, with uppercase A,B,C,D,E and F.
%p Writes the next argument as a pointer address.
%n Writes nothing. Copies instead the number of characters written so far to the location addressed by the next argument.

Further to the above, many additional tweaks exist to the sequences, such as specifying minimum length, signs, etc. More information can be found at sprintf - C++ Reference.

Here are a few examples of calls to the method and their results to illustrate its use.

Table 1.2 sprintf example calls
Format String Arguments Result
"%d" 13 "13"
"+%d degrees" 12 "+12 degrees"
"+%x degrees" 24 "+1c degrees"
"'%c' = 0%o" 65, 65 "'A' = 0101"
"%d * %d%% = %d" 200, 40, 80 "200 * 40% = 80"
"+%d degrees" -5 "+-5 degrees"
"+%u degrees" -5 "+4294967291 degrees"

Hopefully you can already begin to see the usefulness of the function. It does take a fair amount of work to program, but our reward is a very general function we can use for all sorts of purposes.

## 2 Division

Division is the slowest and most complicated of the basic mathematical operators. It is not implemented directly in ARM assembly code because it takes so long to deduce the answer, and so isn't a 'simple' operation.

While this function does look very powerful, it also looks very complicated. The easiest way to deal with its many cases is probably to write functions to deal with some common tasks it has. What would be useful would be a function to generate the string for a signed and an unsigned number in any base. So, how can we go about doing that? Try to devise an algorithm quickly before reading on.

The easiest way is probably the exact way I mentioned in Lesson 1: OK01, which is the division remainder method. The idea is the following:

1. Divide the current value by the base you're working in.
2. Store the remainder.
3. If the new value is not 0, go to 1.
4. Reverse the order of the remainders. This is the answer.

For example:

Table 2.1 Example base 2 conversion
Value New Value Remainder
137 68 1
68 34 0
34 17 0
17 8 1
8 4 0
4 2 0
2 1 0
1 0 1

The unfortunate part about this procedure is that it unavoidably uses division. Therefore, we must first contemplate division in binary.

For a refresher on long division expand the box below.

To implement division in assembly code, we will implement binary long division. We do this because the numbers are stored in binary, which gives us easy access to the all important bit shift operations, and because division in binary is simpler than in any higher base due to the much lower number of cases.

```        1011 r 1
1010)1101111
1010
11111
1010
1011
1010
1```

This example shows how binary long division works. You simply shift the divisor as far right as possible without exceeding the dividend, output a 1 according to the poisition and subtract the number. Whatever remains is the remainder. In this case we show 11011112 ÷ 10102 = 10112 remainder 12. In decimal, 111 ÷ 10 = 11 remainder 1.

Try to implement long division yourself now. You should write a function, DivideU32 which divides r0 by r1, returning the result in r0, and the remainder in r1. Below, we will go through a very efficient implementation.

function DivideU32(r0 is dividend, r1 is divisor)

set shift to 31
set result to 0
while shift ≥ 0

if dividend ≥ (divisor << shift) then

set dividend to dividend - (divisor << shift)
set result to result + 1

end if
set result to result << 1
set shift to shift - 1

loop
return (result, dividend)

end function

This code does achieve what we need, but would not work as assembly code. Our problem comes from the fact that our registers only hold 32 bits, and so the result of divisor << shift may not fit in a register (we call this overflow). This is a real problem. Did your solution have overflow?

Fortunately, an instruction exists called clz or count leading zeros, which counts the number of zeros in the binary representation of a number starting at the top bit. Conveniently, this is exactly the number of times we can shift the register left before overflow occurs. Another optimisation you may spot is that we compute divisor << shift twice each loop. We could improve upon this by shifting the divisor at the beginning, then shifting it down at the end of each loop to avoid any need to shift it elsewhere.

Let's have a look at the assembly code to make further improvements.

.globl DivideU32
DivideU32:
result .req r0
remainder .req r1
shift .req r2
current .req r3

clz shift,r1
lsl current,r1,shift
mov remainder,r0
mov result,#0

divideU32Loop\$:

cmp shift,#0
blt divideU32Return\$
cmp remainder,current

subge remainder,current
sub shift,#1
lsr current,#1
lsl result,#1
b divideU32Loop\$

divideU32Return\$:
.unreq current
mov pc,lr

.unreq result
.unreq remainder
.unreq shift

clz dest,src stores the number of zeros from the top to the first one of register dest to register src

You may, quite rightly, think that this looks quite efficient. It is pretty good, but division is a very expensive operation, and one we may wish to do quite often, so it would be good if we could improve the speed in any way. When looking to optimise code with a loop in it, it is always important to consider how many times the loop must run. In this case, the loop will run a maximum of 31 times for an input of 1. Without making special cases, this could often be improved easily. For example when dividing 1 by 1, no shift is required, yet we shift the divisor to each of the positions above it. This could be improved by simply using the new clz command on the dividend and subtracting this from the shift. In the case of 1 ÷ 1, this means shift would be set to 0, rightly indicating no shift is required. If this causes the shift to be negative, the divisor is bigger than the dividend and so we know the result is 0 remainder the dividend. Another quick check we could make is if the current value is ever 0, then we have a perfect division and can stop looping.

.globl DivideU32
DivideU32:
result .req r0
remainder .req r1
shift .req r2
current .req r3

clz shift,r1
clz r3,r0
subs shift,r3
lsl current,r1,shift
mov remainder,r0
mov result,#0
blt divideU32Return\$

divideU32Loop\$:

cmp remainder,current
blt divideU32LoopContinue\$

subs remainder,current
lsleq result,shift
beq divideU32Return\$

divideU32LoopContinue\$:

subs shift,#1
lsrge current,#1
lslge result,#1
bge divideU32Loop\$

divideU32Return\$:
.unreq current
mov pc,lr

.unreq result
.unreq remainder
.unreq shift

Copy the code above to a file called 'maths.s'.

## 3 Number Strings

Now that we can do division, let's have another look at implementing number to string conversion. The following is pseudo code to convert numbers from registers into strings in up to base 36. By convention, a % b means the remainder of dividing a by b.

function SignedString(r0 is value, r1 is dest, r2 is base)

if value ≥ 0
then return UnsignedString(value, dest, base)
otherwise

if dest > 0 then

setByte(dest, '-')
set dest to dest + 1

end if
return UnsignedString(-value, dest, base) + 1

end if

end function

function UnsignedString(r0 is value, r1 is dest, r2 is base)

set length to 0
do

set (value, rem) to DivideU32(value, base)
if rem > 10
then set rem to rem + '0'
otherwise set rem to rem - 10 + 'a'
if dest > 0
then setByte(dest + length, rem)
set length to length + 1

while value > 0
if dest > 0
then ReverseString(dest, length)
return length

end function

function ReverseString(r0 is string, r1 is length)

set end to string + length - 1
while end > start

setByte(start, temp2)
setByte(end, temp1)
set start to start + 1
set end to end - 1

end while

end function

In a file called 'text.s' implement the above. Remember that if you get stuck, a full solution can be found on the downloads page.

## 4 Format Strings

Let's get back to our string formatting method. Since we're programming our own operating system, we can add or change formatting rules as we please. We may find it useful to add a %b operation that outputs a number in binary, and if you're not using null terminated strings, you may wish to alter the behaviour of %s to take the length of the string from another argument, or from a length prefix if you wish. I will use a null terminator in the example below.

One of the main obstacles to implementing this function is that the number of arguments varies. According to the ABI, additional arguments are pushed onto the stack before calling the method in reverse order. So, for example, if we wish to call our method with 8 parameters; 1,2,3,4,5,6,7 and 8, we would do the following:

1. Set r0 = 5, r1 = 6, r2 = 7, r3 = 8
2. Push {r0,r1,r2,r3}
3. Set r0 = 1, r1 = 2, r2 = 3, r3 = 4
4. Call the function

Now we must decide what arguments our function actually needs. In my case, I used the format string address in r0, the length of the format string in r1, the destination string address in r2, followed by the list of arguments required, starting in r3 and continuing on the stack as above. If you wish to use a null terminated format string, the parameter in r1 can be removed. If you wish to have a maximum buffer length, you could store this in r3. As an additional modification, I think it is useful to alter the function so that if the destination string address is 0, no string is outputted, but an accurate length is still returned, so that the length of a formatted string can be accurately determined.

If you wish to attempt the implementation on your own, try it now. If not, I will first construct the pseudo code for the method, then give the assembly code implementation.

function StringFormat(r0 is format, r1 is formatLength, r2 is dest, ...)

set index to 0
set length to 0
while index < formatLength

if readByte(format + index) = '%' then

set index to index + 1
if readByte(format + index) = '%' then

if dest > 0
then setByte(dest + length, '%')
set length to length + 1

otherwise if readByte(format + index) = 'c' then

if dest > 0
then setByte(dest + length, nextArg)
set length to length + 1

otherwise if readByte(format + index) = 'd' or 'i' then

set length to length + SignedString(nextArg, dest, 10)

otherwise if readByte(format + index) = 'o' then

set length to length + UnsignedString(nextArg, dest, 8)

otherwise if readByte(format + index) = 'u' then

set length to length + UnsignedString(nextArg, dest, 10)

otherwise if readByte(format + index) = 'b' then

set length to length + UnsignedString(nextArg, dest, 2)

otherwise if readByte(format + index) = 'x' then

set length to length + UnsignedString(nextArg, dest, 16)

otherwise if readByte(format + index) = 's' then

set str to nextArg
while getByte(str) != '\0'

if dest > 0
then setByte(dest + length, getByte(str))
set length to length + 1
set str to str + 1

loop

otherwise if readByte(format + index) = 'n' then

setWord(nextArg, length)

end if

otherwise

if dest > 0
then setByte(dest + length, readByte(format + index))
set length to length + 1

end if
set index to index + 1

loop
return length

end function

Although this function is massive, it is quite straightforward. Most of the code goes into checking all the various conditions, the code for each one is simple. Further, all the various unsigned integer cases are the same but for the base, and so can be summarised in assembly. This is given below.

.globl FormatString
FormatString:
format .req r4
formatLength .req r5
dest .req r6
nextArg .req r7
argList .req r8
length .req r9

push {r4,r5,r6,r7,r8,r9,lr}
mov format,r0
mov formatLength,r1
mov dest,r2
mov nextArg,r3
mov length,#0

formatLoop\$:

subs formatLength,#1
movlt r0,length
poplt {r4,r5,r6,r7,r8,r9,pc}

ldrb r0,[format]
teq r0,#'%'
beq formatArg\$

formatChar\$:

teq dest,#0
strneb r0,[dest]
b formatLoop\$

formatArg\$:

subs formatLength,#1
movlt r0,length
poplt {r4,r5,r6,r7,r8,r9,pc}

ldrb r0,[format]
teq r0,#'%'
beq formatChar\$

teq r0,#'c'
moveq r0,nextArg
ldreq nextArg,[argList]
beq formatChar\$

teq r0,#'s'
beq formatString\$

teq r0,#'d'
beq formatSigned\$

teq r0,#'u'
teqne r0,#'x'
teqne r0,#'b'
teqne r0,#'o'
beq formatUnsigned\$

b formatLoop\$

formatString\$:

ldrb r0,[nextArg]
teq r0,#0x0
ldreq nextArg,[argList]
beq formatLoop\$
teq dest,#0
strneb r0,[dest]
b formatString\$

formatSigned\$:

mov r0,nextArg
ldr nextArg,[argList]
mov r1,dest
mov r2,#10
bl SignedString
teq dest,#0
b formatLoop\$

formatUnsigned\$:

teq r0,#'u'
moveq r2,#10
teq r0,#'x'
moveq r2,#16
teq r0,#'b'
moveq r2,#2
teq r0,#'o'
moveq r2,#8

mov r0,nextArg
ldr nextArg,[argList]
mov r1,dest
bl UnsignedString
teq dest,#0
b formatLoop\$

## 5 Convert OS

Feel free to try using this method however you wish. As an example, here is the code to generate a conversion chart from base 10 to binary to hexadecimal to octal and to ASCII.

Delete all code after bl SetGraphicsAddress in 'main.s' and replace it with the following:

mov r4,#0
loop\$:
ldr r0,=format
mov r1,#formatEnd-format
ldr r2,=formatEnd
lsr r3,r4,#4
push {r3}
push {r3}
push {r3}
push {r3}
bl FormatString

mov r1,r0
ldr r0,=formatEnd
mov r2,#0
mov r3,r4

cmp r3,#768-16
subhi r3,#768
cmp r3,#768-16
subhi r3,#768
cmp r3,#768-16
subhi r3,#768

bl DrawString