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    <p>C 2011 (n1570) 6.3.1.8 (“Usual arithmetic conversions”) 1 states that the integer promotions are performed <em>before</em> considering whether the types are the same:</p> <blockquote> <p>Otherwise, the integer promotions are performed on both operands. Then the following rules are applied to the promoted operands:</p> <blockquote> <p>If both operands have the same type, then no further conversion is needed…</p> </blockquote> </blockquote> <p>Thus, in the C abstract machine, <code>unsigned char</code> values must be promoted to <code>int</code> before arithmetic is performed. (There is an exception for perverse machines where <code>unsigned char</code> and <code>int</code> have the same size. In this case, <code>unsigned char</code> values are promoted to <code>unsigned int</code> rather than <code>int</code>. This is esoteric and need not be considered in normal situations.)</p> <p>In the actual machine, operations must be performed in a way that gets the same results <em>as if</em> they were performed in the abstract machine. Because only the results matter, the actual intermediate operations do not need to exactly match the abstract machine.</p> <p>When the sum of two <code>unsigned char</code> values is assigned to an <code>unsigned char</code> object, the sum is converted to a <code>unsigned char</code>. This conversion essentially discards bits beyond the bits that fit in an <code>unsigned char</code>.</p> <p>This means that the C implementation gets the same result whether it does this:</p> <ul> <li>Convert values to <code>int</code>.</li> <li>Add values with <code>int</code> arithmetic.</li> <li>Convert result to <code>unsigned char</code>.</li> </ul> <p>or this:</p> <ul> <li>Add values with <code>unsigned char</code> arithmetic.</li> </ul> <p>Because the result is the same, the C implementation may use either method.</p> <p>For comparison, we can consider this statement instead: <code>int c = a + b;</code>. Also, suppose the compiler does not know the values of <code>a</code> and <code>b</code>. In this case, using <code>unsigned char</code> arithmetic to do the addition could yield a different result than converting the values to <code>int</code> and using <code>int</code> arithmetic. E.g., if <code>a</code> is 250 and <code>b</code> is 200, then their sum as <code>unsigned char</code> values is 194 (250 + 200 % 256), but their sum in <code>int</code> arithmetic is 450. Because there is a difference, the C implementation <strong>must</strong> use instructions that get the correct sum, 450.</p> <p>(If the compiler did know the values of <code>a</code> and <code>b</code> or could otherwise prove that the sum fit in an <code>unsigned char</code>, then the compiler could again use <code>unsigned char</code> arithmetic.)</p>
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    1. POchar and the usual arithmetic conversion rules
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    1. COYou are right. I missed that.
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    2. COWhat you call esoteric (`sizeof(char)=sizeof(int)`) is a real-life scenario for those programming something like 16-bit TI DSPs, where the minimum addressable unit of data memory is 16 bit, so no 8-bit chars.
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