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POWhy are the results of integer promotion different?
 

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  1. POWhy are the results of integer promotion different?
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    <p>Please look at my test code:</p> <pre><code>#include &lt;stdlib.h&gt; #include &lt;stdio.h&gt; #define PRINT_COMPARE_RESULT(a, b) \ if (a &gt; b) { \ printf( #a " &gt; " #b "\n"); \ } \ else if (a &lt; b) { \ printf( #a " &lt; " #b "\n"); \ } \ else { \ printf( #a " = " #b "\n" ); \ } int main() { signed int a = -1; unsigned int b = 2; signed short c = -1; unsigned short d = 2; PRINT_COMPARE_RESULT(a,b); PRINT_COMPARE_RESULT(c,d); return 0; } </code></pre> <p>The result is the following:</p> <pre><code>a &gt; b c &lt; d </code></pre> <p>My platform is Linux, and my gcc version is 4.4.2. I am surprised by the second line of output. The first line of output is caused by integer promotion. But why is the result of the second line different?</p> <p>The following rules are from C99 standard:</p> <blockquote> <p>If both operands have the same type, then no further conversion is needed. Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank.</p> <p>Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type.</p> <p>Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type.</p> <p>Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type.</p> </blockquote> <p>I think both of the two comparisons should belong to the same case, the second case of integer promotion. </p>
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    1. COI think (but I'm not 100% sure) that all integral types shorter than `int` are promoted to `int` for all operations, so `c` and `d` both get genuine promotions and retain their value. `a` is already an `int`, so the comparison *converts* `a` to unsigned.
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    2. COI guessed this result before. But it seems not right according to the C99 standard. I pasted some sections from C99 on my post just now. You could refer to it.
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    3. COThanks for the quote. Perhaps there's another rule somewhere that mentions the initial promotion to `int`?
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