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cbit-manipulationavr

How to get the 8 high bits of a 16-bit variable?

I'm doing some AVR programming and am kinda getting to grips with this whole bit operation thing but still ain't so sure about data type conversion.

For example:

I have a 16-bit variable (myValue) that corresponds the state of 16 LEDs over 2 ports (e.g. 8 LEDs on Port A and 8 on Port D). The 8 high bits of myValue are on Port D.

So my idea was to essentially use (0b0000000011111111 & myValue) to get the values for Port A and (0b1111111100000000 & myValue) for Port D

The question is then can I then do something like uint8_t portA = (0b0000000011111111 & myValue)? I believe I can just simply then set

PORTA |= portA

since Port A has only 8 pins (in my case).

The problem arises in Port D since using unsigned 8 bit on the other hand is not possible for Port D because it will still remain a 16-bit variable after the & operation? How do I then set an 8-pin Port D with a 16-bit variable?

Solution

  • Use:

    PORTD = (myValue >> 8) & 0xff;

    Also stay on the safe side and do the clipping for port A:

    PORTA = myValue & 0xff;

    Be aware that by using |= you can only set bits from 0 to 1, bits that are already 1 will not be altered by the or operation.