Q & AVR MCU Series

their C compiler, instruction set on how to adjust to fit C compiler, a number of suggestions put forward as an example the compiler experts suggest that we brought into place for the immediate SBCI sacrifice ADDI instruction subtraction
Operating instructions for those missing the AVR is easy to prove that the code efficiency should be the realization that we have to find a command instructions and omitted a good way to compromise between
Q: Why did not EORI instruction?
All AVR instructions A few exceptions are two-byte length which means that only 65,536 (64K) possible combinations to arrange the AVR instruction set, instruction set, when we take the specified number of compromises to the fullest possible Unlike the use of these combinations 64K CISC instruction can be a microcontroller, two, three or more bytes of the structure of the AVR instructions we can not achieve all of examples like this as an immediate address register and contains all 32 address 8-bit instructions and registers a constant need to address the need for other five all this instruction will occupy space in the 8K instruction mix in other words we can only be eight such instructions are no more instructions can be achieved while be a length of 17 instructions is not economical and convenient scheme will not be considered in the design process AVR instruction set we listened to a lot of set building meaning we turned to experts in their C compiler, instruction set on how to adjust to fit C compiler, a number of suggestions put forward as an example the compiler experts suggest that we brought into place for the immediate SBCI sacrifice ADDI instruction subtraction
Operating instructions for those missing the AVR is easy to prove that the code efficiency should be the realization that we have to find a command instructions and omitted a good way to compromise between
Q: I use “sbr r30, 3” instruction set in the register 30, bit D3 of the D3 bit does not work but if bit D0 and D1, the practice can be set where I have wrong?
A “SBR” can be used both in a register set number (like “CBR” can be used to clear the register at the same time a number of) command of the second parameter is not specified in the item number but the number of registers used and for logic or operations in this case, the contents of $ 03 and R30 resulting in a minimum-phase or two-bit bit to just set the D3 one should try the following methods
sbr r30 $ 08
sbr r30 0b00001000
sbr r30 (1
The above three lines will have the same results you can choose to use one of your most used
Q: I am multi-tasking system, the interrupt flag in addition to other definitions I need to set and clear the flag should I do?
A register of each symbol with a symbol used to set
ldi flag 01
Or if you want to use the low registers (R0-R15) after reset to clear it and then
inc flag
To set the flag
When the check and clear the flag to do so
lsr flag to move the flag into the flag (flag cleared)
brne flag_was_set If the flag is set to jump
Like you, like all of the interrupt routine before exiting to restore the status register in this way is very safe
Q: How in the AT90S1200 in the constant table function?
A Comparison of the AT90S1200 no LPM instruction is an effective method of constant small table placed in the EEPROM in the address you can use the EEPROM data registers and EEPROM registers effectively, such as direct access to the constant table
. ESEG EEPROM section of the definition
. Db my_var00 = $ 45
. Db my_var01 = $ 4c
. Db my_var02 = $ 5f
. CSEG generic code definitions
If your table is not suitable for placement in the EEPROM in place like this you can in the FLASH
. Def tp = r16 table pointer table index
. Def output = r17 output table in constant
table
ldi output, $ 45
cpi tp, 01
breq end
ldi output, $ 4c
cpi tp, 02
breq end
ldi output, $ 5f
cpi tp, 03
breq end
end
ret
In order to access the table element can be done 2
ldi tp 2
rcall table
When a subroutine return from the table “output” will include “$ 4c”
Q: In the absence of the case ADDI instruction how to register a constant number and the contents immediately add?
A simpler way is a minus a negative number to register the following code is the sum of R20 and 5 in the example of
subi r20 -5
Note that in this case in any case carry flag can not be used to detect if you have the constant overflow into another register, you can use the generic ADD instruction if the constant is 1 you can register using the INC instruction to increase the value of 16-bit The following addition is the same code is r17 r16 and the example of the sum of $ 0b3c
subi r16 low (- $ 0b3c) minus the low byte
sbci r17 high (- $ 0b3c) minus