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    <p>First solutions for shifting (shift is the shift distance, must not be negative, a is the operand to be shifted and contains also the result when done). The power table is used by all three shift operations.</p> <pre class="lang-c prettyprint-override"><code>// table used for shift operations powtab = { 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, -32768 }; // logical shift left if (shift &gt; 15) { a = 0; // if shifting more than 15 bits to the left, value is always zero } else { a *= powtab[shift]; } // logical shift right (unsigned) if (shift &gt; 15) { a = 0; // more than 15, becomes zero } else if (shift &gt; 0) { if (a &lt; 0) { // deal with the sign bit (15) a += -32768; a /= powtab[shift]; a += powtab[15 - shift]; } else { a /= powtab[shift]; } } // arithmetic shift right (signed) if (shift &gt;= 15) { if (a &lt; 0) { a = -1; } else { a = 0; } } else if (shift &gt; 0) { if (a &lt; 0) { // deal with the sign bit a += -32768; a /= powtab[shift]; a -= powtab[15 - shift]; } else { // same as unsigned shift a /= powtab[shift]; } } </code></pre> <p>For AND, OR and XOR i could not come up with a simple solution, so i'll do it with looping over each single bit. There might be a better trick to do this. Pseudocode assumes a and b are input operands, c is the result value, x is the loop counter (each loop must run exactly 16 times):</p> <pre class="lang-c prettyprint-override"><code>// XOR (^) c = 0; for (x = 0; x &lt;= 15; ++x) { c += c; if (a &lt; 0) { if (b &gt;= 0) { c += 1; } } else if (b &lt; 0) { c += 1; } a += a; b += b; } // AND (&amp;) c = 0; for (x = 0; x &lt;= 15; ++x) { c += c; if (a &lt; 0) { if (b &lt; 0) { c += 1; } } a += a; b += b; } // OR (|) c = 0; for (x = 0; x &lt;= 15; ++x) { c += c; if (a &lt; 0) { c += 1; } else if (b &lt; 0) { c += 1; } a += a; b += b; } </code></pre> <p>Thats assuming that all variables are 16 bits and all operations behave as signed (so a&lt;0 actually is true when bit 15 is set).</p> <p>EDIT: i actually tested all possible operand values (-32768 to 32767) for shifts ranging from 0 to 31 for correctness and it works correctly (assuming integer divides). For the AND/OR/XOR code an exhaustive test takes too long on my machine, but since the code for these is pretty simple there should be no edge cases anyway.</p>
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