AVR Bitwise Operations

Question

I am getting started learning more about AVR ATMEGA programming. Reading an article about the inner workings of Arduinos, I am learning how the shiftOut method is structured. I am kind of familiar with bitwise operations so far, but I have an expression I do not understand yet:

void shiftOut(uint8_t dataPin, uint8_t clockPin, uint8_t bitOrder, uint8_t val)
{
 uint8_t i;
 for (i = 0; i < 8; i++) {
 if (bitOrder == LSBFIRST) {
 PORTD |= !!(val & (HIGH << i)); 
 } else {
 PORTD |= !!(val & (HIGH << (7 - i))); 
 } 
 PORTB |= (HIGH << clockPin);
 PORTB ^= (HIGH << clockPin);
 }
}

The line PORTD |= !!(val & (HIGH << i)); is not 100% clear to me. I understand that i set the i-th bit High on PORTD but what das the !! mean and val&(HIGH <<i))

I know this might sound basic, but can you please help me out?

Answer 1

In C, ! is the logical negation operator: !x has the value 1 if x is falsy (i.e. zero), and the value 0 if x is truthy (i.e. non-zero). !!x has just two negations back-to-back, which gives the same logical truth value as x itself, but it collapses all true values to 1. (It's worth remembering that e.g. 2, 0x1000, and -123 are all true in C.)

(val & (HIGH << i))

This is just HIGH shifted left by i bits, and that value anded with val. i.e. if HIGH is 1, it picks the ith bit of val (counting from zero).

So, e.g. if i is 3, (val & (1 << i)) might be 0x08, and the following !! would collapse that to 1.