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    <p>If the number of bits is a compile-time constant:</p> <pre><code>#include &lt;bitset&gt; ... std::bitset&lt;100&gt; b; b[2]=true; </code></pre> <p>If it's not, use <a href="http://www.boost.org/doc/libs/1_40_0/libs/dynamic_bitset/dynamic_bitset.html" rel="noreferrer">Boost.dynamic_bitset</a></p> <p>Or, if you're desperate, std::vector, which is indeed a packed bit vector:</p> <pre><code>#include &lt;vector&gt; ... std::vector&lt;bool&gt; b(100); b[2]=true; </code></pre> <p>You seem to want to use a library that requires bit vectors packed in an array of bytes. Without knowing exactly what order it places the bits in, I can only note that:</p> <p>1) all of the above will probably use at least 32-bit ints with bits ordered least->most or most->least significant</p> <p>2) on little endian (Intel/AMD) CPUs, this means that the memory occupied by the bytes an array of ints may not be consistent with the ordering of bits within the int. if it's "bit 0 is the lsb of int 0, ... bit 32 is the lsb of int 1, ..." then that's the same in little endian as "bit 0 is the lsb of char 0, ... bit 32 is the lsb of char 4 ...", in which case you can just cast a pointer to the int array to a pointer to char array</p> <p>3) supposing the native order of bytes in your bit set / vector isn't exactly what the library needs, then you have to either have to create your own that has the layout they want, or transcribe a copy into their layout. </p> <p>a) if the order of bits within a byte is different, a 256 entry lookup table giving the byte with bits reversed would be efficient. you could generate the table with a small routine.</p> <p>b) to reverse bytes from little&lt;->big endian:</p> <pre><code>inline void endian_swap(unsigned short&amp; x) { x = (x&gt;&gt;8) | (x&lt;&lt;8); } inline void endian_swap(unsigned int&amp; x) { x = (x&gt;&gt;24) | ((x&lt;&lt;8) &amp; 0x00FF0000) | ((x&gt;&gt;8) &amp; 0x0000FF00) | (x&lt;&lt;24); } inline void endian_swap(unsigned long long&amp; x) { x = (x&gt;&gt;56) | ((x&lt;&lt;40) &amp; 0x00FF000000000000) | ((x&lt;&lt;24) &amp; 0x0000FF0000000000) | ((x&lt;&lt;8) &amp; 0x000000FF00000000) | ((x&gt;&gt;8) &amp; 0x00000000FF000000) | ((x&gt;&gt;24) &amp; 0x0000000000FF0000) | ((x&gt;&gt;40) &amp; 0x000000000000FF00) | (x&lt;&lt;56); } </code></pre> <p>To get/set a particular bit within a word, with bit #0 in the least significant bit of word 0:</p> <pre><code>typedef unsigned char block_t; const unsigned block_bits=8; inline void set_bit(block_t *d,unsigned i) { unsigned b=i/block_bits; unsigned bit=i-(block_bits*b); // same as i%b block_t &amp;bl=d[b]; bl|=(1&lt;&lt;bit); // or bit with 1 (others anded w/ 0) } inline void clear_bit(block_t *d,unsigned i) { unsigned b=i/block_bits; unsigned bit=i-(block_bits*b); // same as i%b block_t &amp;bl=d[b]; bl&amp;=(~(1&lt;&lt;bit)); // and bit with 0 (other bits anded w/ 1) } inline void modify_bit(block_t *d,unsigned i,bool val) { if (val) set_bit(d,i) else clear_bit(d,i); } inline bool get_bit(block_t const* d,unsigned i) { unsigned b=i/block_bits; unsigned bit=i-(block_bits*b); // same as i%b return d[b]&amp;(1&lt;&lt;bit); } </code></pre> <p>Obviously if the rule for bit organization differs, you have to change the above.</p> <p>Using the widest possible int your CPU processes efficiently as block_t is best (dont' forget to change <code>block_bits</code>), unless the endianness doesn't work out w/ the library you're using.</p>
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    1. POHow to read and write bits to a byte array
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    1. USJonathan Graehl
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    1. COThanks so much! That's brilliant (would +1 but need 15 rep). If you don't mind, please would you tell me how I would write a short and or character using your set_bit functions (don't waste time padding, I know how to do that :))
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