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copied!<p>If you take a trivial example of <code>4 (0b0100) + 5 (0b0101)</code> you can see that the unsigned sum should be <code>9 (1001)</code> which is actually <code>-7</code> in two's complement. If you then take that sum (0b1001) and subtract 4 from it using two's complement arithmetic: </p> <pre><code> 0b1001 - 0b0100 = 0b1001 + 2s_complement(0b0100) = 0b1001 + 0b1100 = 0b1_0101 </code></pre> <p>you end up with 0101 which is 5 (you drop the overflowing most significant 1 during a 2's complement operation). Subtracting 5 from the sum equals 4:</p> <pre><code> 0b1001 - 0b0101 = 0b1001 + 2s_complement(0b0101) = 0b1001 + 0b1011 = 0b1_0100 </code></pre> <p>This satisfies the c code you provided but still resulted in an overflow.</p> <p>From wikipedia's article on <a href="http://en.wikipedia.org/wiki/Two%27s_complement" rel="nofollow">two's complement</a>:</p> <pre> Two's complement Decimal 0111 7 0110 6 0101 5 0100 4 0011 3 0010 2 0001 1 0000 0 1111 −1 1110 −2 1101 −3 1100 −4 1011 −5 1010 −6 1001 −7 1000 −8 </pre> <p><strong>Update:</strong><br> To demonstrate your INT_MAX example using my trivial 4 bit integer system with INT_MAX = 7 we can see the same result as your c code. </p> <pre><code> 7 + 7 (0b0111 + 0b0111) = 0b1110 (-2 in two's complement) </code></pre> <p>Just like my example above, subtracting, <code>sum - 7</code> will equal <code>7</code>. </p> <pre><code> 0b1110 - 0b0111 = 0b1110 + 2s_complement(0b0111) = 0b1110 + 0b1001 = 0b1_0111 </code></pre>

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