c - Bitwise shifting array of char's

Question

I have got an array of chars that I'm trying to bitwise shift right >>, then & with another array. I think I have got the wrong idea of how to do this.

I thought, even though it was an array of chars just stating my_array >>= 1 would shift everything but I am getting an error: "error: invalid operands to binary >> (have ‘char[8]’ and ‘int’)"

The bitwise comparision I am trying to do is with a similar size array initiated to all "0's"...for that I'm getting: "error: invalid operands to binary & (have ‘char *’ and ‘char *’)"

Do I need to convert these array's into something else before I can shift and compare?

Sorry, I was not super clear... All great advice up to this point and I think I am realizing more that there is no super easy way to do this. More specifically, what I am trying to do is shift the bits of the WHOLE char array right 1, adding the bit shifted off the right back to the left most side of the array, do the bitwise compare with another array of same size.

Technically the compare doesn't have to be array with array... I just need the bits. Would it be easier to convert the array's to something else before trying to do the shifts/comparisons?

Answer 1

You have to shift and compare elementwise.

for(i = 0; i < len; ++i)
    array[i] >>= 3;

for example. If you want to move the bits shifted out of one element to the next, it's more complicated, say you're shifting right, then

unsigned char bits1 = 0, bits2 = 0;
for(i = len-1; i >= 0; --i) {
    bits2 = array[i] & 0x07;
    array[i] >>= 3;
    array[i] |= bits1 << 5;
    bits1 = bits2;
}

traversing the array in the other direction because you need the bits from the next higher slot.