bit comparison in loop on AVRs

Question

I'm learning about bit logic in C on AVRs and I have a problem.

I want to compare an "i" bit (from the right) from int8_t variable and if it is 1, then do the next instruction, but it doesn't work. Here's what I write:

if (variable & (1<<i)==(1<<i)) instruction;

In example for following data:

uint8_t dot=0101;
PORTC=1;
for (int i=0; i<4; i++)
{
    PORTB = fourDigit[i];
    if (dot & (1<<i)==(1<<i)) PORTB--;
    PORTC<<=1;
}

The dot (as it is connected to PB0) should illuminate on the first and third digit, but at present it lamps on every digit. What's the problem?

Thanks for your time.

score 1 · Accepted Answer · answered May 27 '15 at 21:22

It is done by bit masking. If you want to check whether or not an i'th bit of a is 1 you will do something like this:

if (a & (1 << i))
{
    // Do something
}

This way all of the bits of a except the i'th one will be ANDed with zeros, thus getting a value of zero. The i'th bit will be ANDed with 1, thus not changing it's value. So the if condition will be true in case the bit is not zero, and false otherwise.
The comparison code you are presenting should work as well, but I suspect the dot variable is not containing the value you think it is containing. uint8_t dot=0101; makes it to be equal to 101 in octal base (due to the leading zero) or 65 in decimal. Not 101 in binary.

I've copied your code and it works. My should work as well, but it didn't. As to the number inside uin8_t dot, I've used it in binary (0b0101). Thank you — Lexander, May 27 '15 at 21:34
Yep. The precedence is the case here. You should enclose the left part of the `==` operator in brackets. — Eugene Sh., May 27 '15 at 21:37

bit comparison in loop on AVRs

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