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Current direction:

Start with and unsigned char which is 1 Byte on my system using sizeof. Range is 0-255. If length is the number of bits I need then elements is the number of elements (bytes) I need in my array.

constant unsigned int elements = length/8 + (length % y > 0 ? 1 : 0);  
unsigned char bit_arr[elements];

Now I add basic functionality such as set, unset, and test. Where j is the bit per byte index, i is the byte index and h = bit index. We have i = h / 8 and j = i % 8.

Psuedo-Code : 

bit_arr[i] |= (1 << j); // Set 
bit_arr[i] &= ~(1 << j);  // Unset
if( bit_arr[i] & (1 << j) ) // Test
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    Avoid "pow". Use bit-shifting << and >> operators instead. (1 << i) == 2^i. Commented May 13, 2011 at 23:08
  • I did not know you can use bit shifting to do that. Thank you. Commented May 13, 2011 at 23:14
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    <pedantry>There's nothing that guarantees a byte == 8 bits. It can differ per platform/CPU. IIRC CHAR_BITS will tell you how many bits are in a char (which is the same as a byte in C, BTW -- it's defined as the smallest individually addressable unit of memory). </pedantry> Commented May 13, 2011 at 23:15
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    if( bit_arr[i] &= pow(2,j) ) for the test is wrong, just use & not &=. Commented May 13, 2011 at 23:19

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Looks like you have a very good idea of what needs to be done. Though instead of pow(2, j), use 1 << j. You also need to change your test code. You don't want the test to do an assignment to the array.

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pow() will give you floating-point values, which you don't want. At all. It might work for you, as you use powers of two, but it can get weird as j gets bigger.

You'd do a bit better to use 1 << j instead. Removes any chance of float weirdness, and it probably performs better, too.

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