Consider this AVR assembly code:
blink:
sbi PINA, 0 ; Toggle PINA
jmp blink
According to the documentation, this should set bit 0 on port A, however, this code actually causes the bit to toggle.
Why is that?
CodePudding user response：
Why is that?
Because this is how it works on newer AVR devices: You can toggle a port pin by SBI-ing the PIN register. An example is ATmega88, whereas older ATmega8 just sets a bit in PIN.
Advantage is that this allows to atomically toggle a port pin, whereas the traditional IN-XOR-OUT is not atomic, slower and consumes more flash. To make it atomic, you'd have to (temporarily) disable IRQs.
Cf. for the example data sheet of ATmega48/88/168, Section 13 I/O-Ports:
13.2.2 Toggling the Pin
Writing a logic one to PINxn toggles the value of PORTxn, independent on the value of DDRxn. Note that the SBI instruction can be used to toggle one single bit in a port.
CodePudding user response：
Why? Because hardware wiring will do it. Let look how IO pin is realized.
Here you can see logic diagram for every pin. Output value is driven by D flip-flops in PORTA register. DDRA register enable outputting value to pin. And PINA is second D flip flop in synchronizer. A read the PINA reads value from this flip-flop. As you can see write to PIN (WPx signal) is connected to multiplexor which select input signal for PORTA output flip-flop. In case writing one this mux is switched to upper signal where is connected inverted state of outputting value. Also clk pulse for PORTA flip-flop is generate. Result is that PORTA change his logical value.
sbi PINA, 0 instruction does not toggle PINA but toggle PORTA. PINA always follow signal on the pin after one CPU clock cycle.
If you want set bit 0 of the PORTA, you must use
SBI PORTA, 0