# tianjara.net | Andrew Harvey's Blog

### Entries tagged "avr".

8th May 2009

Just quickly because I need to get back to work, here is how I managed to finally set up a working environment to use the COMP2121 AVR Board at home using linux.

1. Installed Sun's Virtual Box.
2. Downloaded and installed Windows 7 Beta (Because its free (as in free beer) and only expires after this course is over)
3. Under the network settings for the Windows 7 virtual machine, disabled it from the rest of the world (I don't need it here, so save my download for better things that Windows updates).
4. Install AVR Studio.
5. Install a driver (I used this one http://www.ftdichip.com/Drivers/CDM/CDM%202.04.16.exe) from here http://www.ftdichip.com/Drivers/VCP.htm. Don't need to worry about trusting anything as this virtual will just be used for the AVR board.
6. In the bottom right hand of the virtual box window enable the "FTDI USB <->UNSW uLab" USB device (when its plugged in).
7. Run nite.exe with these arguments " -l com3 -f 4 -t 1 -h X:\AVRProj\". You may need to check if its com3 or something else in "Device Manager". Where that last path is the default directory (I set up a shared directory in the VM settings).
Tags: avr, comp2121.
17th April 2009

[Updated: Clarified what I wrote about the way stacks in AVR work, in the Push/Pop section.]

## Data Transfer Instructions

(Credit: Hui Wu/COMP2121 Lecture Notes)

ld Rd, v
Rd $\in$ {r0, r1, ..., r31} and v $\in$ {x, x+, -x, y, y+, -y, z, z+, -z}

(remember the X, Y, Z pointers)

Load Program Memory (lpm):

Can take on three forms,

Syntax Operation
lpm
R0⟵(Z)
lpm Rd, Z
R0⟵(Z)
lpm Rd, Z+
Rd⟵(Z)
Z⟵Z+1

Z contains the byte address while the program flash memory uses word addressing. Therefore the word address must be converted into a byte address before having access to the data on the program flash memory. Example usage, (Table_1<<1 converts the word address into a byte address)

   ldi zh, high(Table_1<<1) ; Initialise Z pointer
ldi zl, low(Table_1<<1)
lpm r16, z+ ; r16=0x76
lpm r17, z ; r17=0x58
...
Table_1: .dw 0x5876
..

IN/OUT:

AVR has 64 IO registers. Example,

in Rd, A
out A, Rd
where 0 ≤ A ≤ 63.

Push/Pop:

The stack pointer (implemented as two 8-bit registers in the I/O space) points to the next free space in RAM above/below (depends how you look at it) the stack. The stack in AVR is part of the SRAM space, and the stack (in AVR) grows from higher memory locations to lower memory locations. I'm talking about the actual value of the address, so 0xFFFF is a higher address than 0xDDDD. This got me a little confused at one stage because if you draw a picture of memory with 0x0000 at the top of the diagram, and 0x0001 below it and so on then in reference to the diagram a stack that is getting larger with PUSH operations is growing upwards (you usually associate higher to lower with down, this is why I got confused).

So the first thing you must do is initialise the stack pointer (RAMEND is a good starting location).

So a push operation,

push Rr

will push Rr onto the stack (ie. put the contents of Rr into the location that the SP points to), and then decrement the SP by 1. Pop has a similar opposite effect.

## Shift Instructions

### Logical shift left

lsl Rd

### Rotate Left Through Carry

rol Rd

Both operation change some status register flags.

## Functions

We dabbed into this in first year but just to revise and extend a little I'll try to reiterate this here.

The heap is used for dynamic memory applications (eg. malloc()).

The stack is used to store return addresses, parameters, conflict registers and local variables and other things.

In passing parameters in WINAVR (C compiler for AVR) for say a function call they are passed by value for scalar variables (eg. char, int, float) and passed by reference for non-scalar variables (eg. array, struct).

Rules are needed between the caller and the callee to resolve issues such as,

• how to pass values and references to a function?
• where to get the return value?
• how to handle register conflicts? (if a function wants to use a register that was previously in use)
• how to allocate and deallocate stack memory to and from local variables?

If a register is used in both caller and callee and the caller needs its old value after the return from the callee, then a register conflict occurs. Either the compiler or the assembly programmer needs to check for this. The work around is to save the conflict registers on the stack.

The return value of a function needs to be stored in designated registers. WINAVR uses r25:r24 for this.

A stack consists of stack frames. A stack frame is created whenever a function is called, and it is freed whenever the function returns.

(Credit: Hui Wu/COMP2121 Lecture Notes)

## Macros

The AVR assembler offers macros. A macro is just a segment of code that you define and can then use by just calling the macro. Basically the macro name is just a place holder for the macro code. When the program is assembled the macro name will be replaced by the code that macro defines. This defines a macro named mymacro,

.macro mymacro
lds r2, @0
lds r3, @1
sts @1, r2
sts @0, r3
.endmacro

We can then invoke this macro with,

mymacro p, q

The p, q are used like arguments. So @0 will be replaced with p and @1 will be replaced by q. In AVR you can used @0 up to @9 in the macro body.

## Assembly Process

The AVR assembler uses a two pass process.

Pass One:

• Lexical and syntax analysis: checking for syntax errors.
• Record all symbols (labels...) in a symbol table.
• Expand macro calls.

Pass Two:

• Use symbol table to substitute the values for the symbols and evaluate functions (eg. low(-13167)).
• Assemble each instruction (ie. generate machine code). For example add Rd, Rr is encoded as machine code as 0000 11rd dddd rrrr.

## References

Wu, Hui. COMP2121 09s1 Lecture Notes.

Tags: avr, comp2121, computing.
14th April 2009

Just some notes on using ADD and ADC (and also a not on compare and branch instructions at the end). (Thanks also to my lab partner who helped me with some of this stuff.)

The registers for AVR are 8 bits long, as noted if I want to store a 16 bit number I need to use two registers, one will store the low byte and the other will store the high byte. eg the 16 bit binary number a = 1111111100000000 (as big endian) would be stored as a_high = 11111111, a_low = 00000000.

Another thing that I've learnt is how addition of multi byte numbers in AVR works. AVR has the instructions ADD (add without carry) and ADC (add with carry). Lets look at a simple case first.

ldi r16, 0b10101010 //the 0b indicates binary (\$ or 0x for hex, nothing for decimal, 0 for octal). loads the binary number 10101010 into register 16.
ldi r17, 0b10101010
add r16, r17

What happens here is,

  10101010
+ 10101010
01010100
c c c c

The carry bit is set, and hence and overflow has occurred. The actual answer is 101010100 but this has 9 bits and cannot fix in the 8 bits of space we have for the result to be stored into r16. Here we used the ADD instruction this means (at least to the best of my understanding, please correct me if I'm wrong) that we ignore what is originally in the carry flag (part of the status register). If we used ADC and the carry flag was set to 1 then when we add the right most bit as above 0+0 = 0 but we have a carry so we would get a 1 in that last bit. We can use this to do multi byte arithmetic.

Say we have,

//a is a 16 bit number 1010101010101010
ldi al, 0b10101010
ldi ah, 0b10101010
//b is a 16 bit number 1010101010101010
ldi bl, 0b10101010
ldi bh, 0b10101010

//to do a+b (and store the result in a) we do
adc ah, bh

We can then extend this to as many bytes as we want. Say we have a 3 byte number spread over the registers of a_0, a_1, a_2 and another 3 byte number in the registers b_0, b_1, b_2. We could add them like this,

add a_0, b_0
adc a_2, b_2

The result would be spread over a_0, a_1 and a_2 (remember though that we can still overflow this).

It was also enlightening (though also a nice reminder) to see how this low level addition works. (when doing 1bit addition (without carry), a + b, the sum = a xor b and the carry = a and b.) (see picture below)

(Credit: Annie Guo, Basics of Digital Systems, COMP2121 Notes)

This concept can similarly be applied to other instructions. For example,

cp a, b
breq label

will compare a and be and branch to the label label if a is equal to b (and if not continue on executing the subsequent instructions). However what if you wanted to compare if a two byte number was equal to a two byte number? Sure you could just repeat the code one time for the low byte and another for the high byte but you can also do this,

cp al, bl
cpc ah, bh
breq label

These cp and cpc instructions use some trickery (well not really but I didn't initially see how this worked, though its quite obvious now) where they modify some flags based on the result of the comparison and then the branch instructions (breq, brne, brlo...) look at the respective flag to see if they should branch or not.

Tags: avr, comp2121, computing.