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    <p>I think you are confusing two things. <em>Promotion</em> is the process by which values of integer type "smaller" that int/unsigned int are converted either to int or unsigned int. The rules are expressed somewhat strangely (mostly for the benefit of handling adequately char) but ensure that value and sign are conserved.</p> <p>Then there is the different concept of <em>usual arithmetic conversion</em> by which operands of arithmetic operators are converted to a common type. It begins by promoting the operand (to either int or unsigned) if they are of a type smaller than int and then choosing a target type by the following process (for integer types, 6.3.1.8/1)</p> <blockquote> <p>If both operands have the same type, then no further conversion is needed.</p> <p>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>(Note that ISTR that those rules have changed slightly between C89 and C99)</p>
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    1. POIn a C expression where unsigned int and signed int are present, which type will be promoted to what type?
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    1. USAProgrammer
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    1. COC99 made an explicit use of the term "rank" of an integral type. Also, see my comment to this answer: http://stackoverflow.com/questions/2032744/unexpected-behavior-when-printing-4-byte-integer-byte-by-byte/2032769#2032769
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