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In this assignment you’re tasked with creating more functions using assembly language. There are two big differences between this assignment and the last:
As before, you will likely want to keep the AVR Assembly Reference handy as well as the rules for register usage. Make sure you are using registers appropriately - we will be checking for this when you demo. Some of the functions you have to complete are both called and call another function — they need to follow both the caller-saved and the callee-saved conventions!!! Use the stack!
The goal of this assignment is to implement basic functions in assembly language. The functions chosen are examples that may actually be implemented in assembly rather than a higher level language.
The main concepts are:
Step 1: 9-4 = 5 Step 2: 5-4 = 1 We have less than 4 left, so we're done. There were just 2 steps, so 9/4 must be 2.
Another example. 36/7:
Step 1: 36-7 = 29 Step 2: 29-7 = 22 Step 3: 22-7 = 15 Step 4: 15-7 = 8 Step 5: 8-7 = 1 We have less than 7 left, so we're done. There were just 5 steps, so 36/7 must be 5.
(This, of course, isn’t the approach used by “long division” that you learned in school. Using the Big-O notation covered in CSE 247, the approach shown here does O(n) steps. Typically computers use a different approach that is quite like “long division” and only does O(log n) steps, which is much faster for large enough n. Consequently we’re referring to the approach here as “slow”).
Modulus: Modulus is related to division however instead of the quotient, we are looking for the remainder. For this problem, we want you to use the division function you wrote (by calling it from your modulus function), taking the result and performing multiplication (using the
mul AVR Instruction, NOT the method you wrote in studio), then finally some subtraction to determine the remainder. Note again, that this is not the most efficient way of performing modulus, but it does give you good practice using various AVR instructions.
To keep things as simple as possible all values are unsigned. In C the 8-bit values are
uint8_t (which is the same as a
byte and 16-bit values are
In your repository you will find an
assignment09 Arduino project. This project contains the prototypes for the functions we are asking you to write as well as code to test your work.
The Assembly language functions may be easier to write if you have already tested the logic using C. There are four C-functions you’ll need to complete in
uint8_t slowDivisionAlgorithm8(uint8_t dividend, uint8_t divisor): This function will do division of 8-bit values the “slow” way.
uint16_t slowDivisionAlgorithm16(uint16_t dividend, uint16_t divisor): This function will do division of 16-bit values the “slow” way. (And may look a lot like the 8-bit version. None the less, note the differences!)
uint8_t slowModulusAlgorithm(uint8_t dividend, uint8_t divisor): This function will return the remainder of the dividend divided by the divisor, and should call your
uint16_t sumOdds(uint8_t a, uint8_t b): This function will sum all of the odd values that occur between a and b (inclusive). You should use your
modulusfunction to solve this problem.
There are five functions to complete entirely in assembly language:
uint8_t slowDivisionUint8(uint8_t a, uint8_t b): An 8-bit division using assembly language. Use the C-code version as a guide for the logic of your assembly language version.
bool greaterThanOrEqualUInt16(uint16_t a, uint16_t b): Testing logic conditions on multi-byte values is a little trickier than just the
cpbased comparisons we’ve covered so far. Create a multi-byte comparison function. It may help to think carefully about a variety of test cases (and even write them out as 2-byte hex values).
uint16_t slowDivisionUint16(uint16_t a, uint16_t b): Implement a 16-bit division algorithm. Again, use your C-code version as a prototype. (Hint: You may need to do comparisons of 16-bit values. You may want to try to
callon some of your earlier work)
unit16_t slowModulusUint8(unit8_t a, uint8_t b): Implement a modulus operation on 8-bit operators, using your
slowDivisionUint8function. To ensure that you’re using proper caller-saved and callee-saved register usage for this function, you should be able to seamlessly replace calls to your assembly function
slowDivisionUint8with calls to your corresponding C function
slowDivisionAlgorithm8and get the same result. TAs will check for this when you demo.
uint16_t sumOddsUint8(uint8_t a, uint8_t b): This function will sum all of the odd values that occur between a and b (inclusive). You should use your
slowModulusUint8function to solve this problem. To ensure that you’re using proper caller-saved and callee-saved register usage for this function, you should be able to seamlessly replace calls to your assembly function
slowModulusUint8with calls to your corresponding C function
slowModulusAlgorithmand get the same result. TAs will check for this when you demo.
Assembly language can be challenging. The provided files include several test cases to help you with your work. Here’s the approach we suggest:
setup()includes calls to all test cases. There are two styles of tests:
Standalone test cases, like:
uint8_t dividend8 = 175; uint8_t divisor8 = 26; uint8_t quotient8ASM = slowDivisionUint8(dividend8,divisor8); uint8_t quotient8Alg = slowDivisionAlgorithm8(dividend8,divisor8); pprintf("slowDivisionAlgorithm8(%u,%u) is %u (algorithm) and %u (assembly); Should be %u\n", dividend8, divisor8, quotient8Alg, quotient8ASM, dividend8/divisor8);
These tests will run just a single test case, which is probably what you need to do when debugging.
More rigorous tests are included in “test functions”, like on Assignment 10. They can be run via function call, like:
// Full Test: testSlowDivisionUint8();
You will want to comment/un-comment test cases so you can focus on each part independently.
It’s probably best start with the C-code version of 8-bit division. I.e., complete
slowDivisionAlgorithm8() in assignment09.ino. Use the test functions to test it (they will indicate the values printed by both the algorithm, which should be correct when you’re done with the C-code, and the assembly language, which will still be incorrect)
Write and test the assembly language version of 8-bit division:
slowDivisionUint8 in assignment09.S.
Try the assembly language version of comparisons — do
greaterThanOrEqualUInt16 in assignment09.S. Again, test your work.
Do the C version of 16-bit division. Complete
slowDivisionAlgorithm16() in assignment09.ino.
Now try the assembly language version of 16-bit division:
slowDivisionUint16 in assignment09.S. This may be the most difficult part of the assignment. (Hint: 16-bit subtraction can be done with just two instructions)
Complete the two modulus functions:
slowModulusUint8. Don’t forget to use your division functions in your solution - we will be checking for this.
If you do not get the message
Done with tests!!! at the end of your testing functions, you did not successfully finish all tests. You are probably caught in an infinite loop somewhere…
If you call another function from within the function you are writing, keep in mind that the two functions may use the same registers. Be careful not to overwrite data you may need later. Convention is to save all caller-saved registers before calling another function. This can be done using the
Pay attention to which test cases are failing. For example, does
sumOddsUInt8 only fail when the minimum value is even? Think about the actual binary of the bytes stored in each register; it can help guide you to a solution.
It is in your best interest to read up on the AVR Multiplication Instruction before attempting to complete the modulus function.
Note that the multiplication instruction is expecting two unsigned 8-bit values as input, producing a 16 bit result. This does not quite match up with what we will have since one of our values will be an 8-bit unsigned integer (the divisor), but one of them will be a 16-bit unsigned integer (the output from the
slowDivision function). For the purposes of this function, you may simply ignore the most significant byte of the 16-bit value and perform the multiplication with the least significant byte.
The output of the multiplication instruction will be placed in
r1. Note, however, that
r1 is a special register that is expected to contain zero. What this means is that if you use a multiplication instruction, you must read the value out of
r1 and set it back to zero as soon as possible.
slowDivisionAlgorithm8()in place of
slowModulusAlgorithm()in place of
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