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cbit-fields

Pushing boolean values of an array to uint8 bitfield

Let's suppose I have:

myArray[7] = {TRUE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE}

uint8 mybitfield;

What is the most efficient way to "push" those values to an uint8 bitfield with 0=FALSE, 1=TRUE

So that mybitfield is represented as: 

[1,0,0,1,1,0,0,0]

(The Least Significant Bit is not considered and will be always 0).

Thanks!

Solution

  • As already noted, you have to iterate over the bits individually, for example:

    int myArray[7] = {TRUE,FALSE,FALSE,TRUE,TRUE,FALSE,FALSE};

uint8_t bitfield = 0;
for (int i = 0; i < 7; ++i) {
    bitfield |= myArray[i] ? 1 : 0;
    bitfield <<= 1;
}

    This results in 0b10011000, i.e., the array has the most significant bit first and an implicit zero for the least significant bit.