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    <p>Signed integer variables do not have wrap-around behavior in C language. Signed integer overflow during arithmetic computations produces <em>undefined behavior</em>. Note BTW that GCC compiler you mentioned is known for implementing <em>strict overflow semantics</em> in optimizations, meaning that it takes advantage of the freedom provided by such undefined behavior situations: GCC compiler assumes that signed integer values never wrap around. That means that GCC actually happens to be one of the compilers in which you <em>cannot</em> rely on wrap-around behavior of signed integer types.</p> <p>For example, GCC compiler can assume that for variable <code>int i</code> the following condition</p> <pre><code>if (i &gt; 0 &amp;&amp; i + 1 &gt; 0) </code></pre> <p>is equivalent to a mere</p> <pre><code>if (i &gt; 0) </code></pre> <p>This is exactly what <em>strict overflow semantics</em> means.</p> <p>Unsigned integer types implement modulo arithmetic. The modulo is equal <code>2^N</code> where <code>N</code> is the number of bits in the value representation of the type. For this reason unsigned integer types do indeed appear to wrap around on overflow.</p> <p>However, C language never performs arithmetic computations in domains smaller than that of <code>int</code>/<code>unsigned int</code>. Type <code>unsigned short int</code> that you mention in your question will typically be promoted to type <code>int</code> in expressions before any computations begin (assuming that the range of <code>unsigned short</code> fits into the range of <code>int</code>). Which means that 1) the computations with <code>unsigned short int</code> will be preformed in the domain of <code>int</code>, with overflow happening when <code>int</code> overflows, 2) overflow during such computations will lead to undefined behavior, not to wrap-around behavior.</p> <p>For example, this code produces a wrap around</p> <pre><code>unsigned i = USHRT_MAX; i *= INT_MAX; /* &lt;- unsigned arithmetic, overflows, wraps around */ </code></pre> <p>while this code</p> <pre><code>unsigned short i = USHRT_MAX; i *= INT_MAX; /* &lt;- signed arithmetic, overflows, produces undefined behavior */ </code></pre> <p>leads to undefined behavior.</p> <p>If no <code>int</code> overflow happens and the result is converted back to an <code>unsigned short int</code> type, it is again reduced by modulo <code>2^N</code>, which will appear as if the value has wrapped around.</p>
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    1. COwhat does "unsigned i" stand for? You've explained it very well, could you show it on this example: `short int y = 511, z = 512; y*=z;`
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    2. CO@orustammanapov: `unsigned i` stands for `unsigned int i`, just like `long i` stands for `long int i`. `int` is implied in such contexts. As for your example, `y*=z` is interpreted as `y = (short) ((int) y * (int) z)`, which boils down to `y = (short) 261632`. Note that in this case there's no overflow during arithmetic evaluations (which fit into `int` perfectly well), but there is overflow during conversion back to `short`. The behavior in this case is *implementation-dependent*.
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