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    <div class="moz-cite-prefix">Yes this is what I think I'm after....<br>
      <br>
      <b><a href="http://en.wikipedia.org/wiki/Bit_array">http://en.wikipedia.org/wiki/Bit_array</a><br>
        <br>
        <br>
      </b>On 18/11/14 17:44, Rick Forrister wrote:<br>
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    <blockquote cite="mid:546AEAD7.7070303@charter.net" type="cite">
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      In C++ I found the following code sample:<br>
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              <div class="vote"> </div>
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                <p>Nobody mentioned the Standard C++ Library: <a
                    moz-do-not-send="true"
                    href="http://www.sgi.com/tech/stl/bitset.html"><code>std::bitset<N></code></a>.<br>
                  Or the boost version: <a moz-do-not-send="true"
href="http://www.boost.org/doc/libs/1_36_0/libs/dynamic_bitset/dynamic_bitset.html"><code>boost::dynamic_bitset</code></a>.</p>
                <p>No need to roll your own:</p>
                <pre style="" class="default prettyprint prettyprinted"><code><span class="com">#include</span><span class="pln"> </span><span class="str"><bitset></span><span class="pln">
</span><span class="com">#include</span><span class="pln"> </span><span class="str"><iostream></span><span class="pln">

</span><span class="kwd">int</span><span class="pln"> main</span><span class="pun">()</span><span class="pln">
</span><span class="pun">{</span><span class="pln">
    std</span><span class="pun">::</span><span class="pln">bitset</span><span class="pun"><</span><span class="lit">5</span><span class="pun">></span><span class="pln"> x</span><span class="pun">;</span><span class="pln">

    x</span><span class="pun">[</span><span class="lit">1</span><span class="pun">]</span><span class="pln"> </span><span class="pun">=</span><span class="pln"> </span><span class="lit">1</span><span class="pun">;</span><span class="pln">
    x</span><span class="pun">[</span><span class="lit">2</span><span class="pun">]</span><span class="pln"> </span><span class="pun">=</span><span class="pln"> </span><span class="lit">0</span><span class="pun">;</span><span class="pln">
    </span><span class="com">// Note x[0-4]  valid</span><span class="pln">

    std</span><span class="pun">::</span><span class="pln">cout </span><span class="pun"><<</span><span class="pln"> x </span><span class="pun"><<</span><span class="pln"> std</span><span class="pun">::</span><span class="pln">endl</span><span class="pun">;</span><span class="pln">
</span><span class="pun">}</span></code></pre>
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      and you could map them more or less as follows:<br>
      sex=m|f=0|1<br>
      mood=h|s=0|1<br>
      mature=a|c=0|1<br>
      color=r|b=0|1<br>
      handed=r|l=0|1<br>
      Thus, a female child who is happy, likes re and is right handed
      would encode to: 10100<br>
      <br>
      in PDP-11 assembler there's a BSB command (bit set byte), and you
      can embed assembler in C functions.<br>
      <br>
      Hope it helps<br>
      RickF<br>
      <br>
      <div class="moz-cite-prefix">On 11/17/2014 9:28 PM, James McDonald
        wrote:<br>
      </div>
      <blockquote cite="mid:546AD914.9080106@jamesmcdonald.id.au"
        type="cite">Hi All, <br>
        <br>
        I am wondering how to do the following: <br>
        <br>
        Say I have a list of values I want to assign to a person e.g. <br>
        <br>
        1. Male <br>
        2. Female <br>
        3. Happy <br>
        4. Sad <br>
        5. Child <br>
        6. Red is favorite color <br>
        7. Blue is favorite color <br>
        8. RIght handed <br>
        9. Left handed <br>
        etc. <br>
        <br>
        Now how do I make the list contain numerical values that I can
        then add together and then end up with a number that I can
        reverse and discern which values were assigned to the person? <br>
        <br>
        I believe that there is something in binary that means you can
        assign bit or a corresponding number to each option and then
        pull out the selections programmatically <br>
        <br>
        My problem is I can't figure out what to Google to teach myself
        how to do it. <br>
        <br>
        Can anyone point me in the right direction? <br>
        <br>
        Thanks. <br>
        James McD. <br>
        <br>
        <br>
        <br>
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