Christian Böhme wrote: > Andrew Haley wrote: > >> Section 5 Para. 9 > > That's talking about _relational_ operators and starts with > > "The usual arithmetic conversions are performed on operands of > arithmetic or enumeration type." No, it isn't. ISO/IEC 14882:1998(E) 5 Expressions 9 Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called the usual arithmetic conversions, which are defined as follows: - If either operand is of type long double, the other shall be converted to long double. - Otherwise, if either operand is double, the other shall be converted to double. - Otherwise, if either operand is float, the other shall be converted to float. - Otherwise, the integral promotions (4.5) shall be performed on both operands.54) - Then, if either operand is unsigned long the other shall be converted to unsigned long. - Otherwise, if one operand is a long int and the other unsigned int, then if a long int can represent all the values of an unsigned int, the unsigned int shall be converted to a long int; otherwise both operands shall be converted to unsigned long int. - Otherwise, if either operand is long, the other shall be converted to long. - Otherwise, if either operand is unsigned, the other shall be converted to unsigned. [Note: otherwise, the only remaining case is that both operands are int ] > going on about pointer conversions ectpp. My example did, > however, neither. Any more pointers ? > >> and Section 4.5. > > That's "Integral promotions" Well, yes. That's what applies here. The integral promotions are relevant, because they are *directly* referred to in the paragraph above. > vs. 4.7 "Integral conversions" as in the subject of the OP. Right, and you have now had the exact language quoted to you. Your first example was int64_t a; uint32_t b = 8; a = -(b * 2u); So, the last section of Para 9 applies: b is promoted to unsigned int, and the result, and unsigned int, is negated, yielding another unsigned int. In this case: int64_t a; uint32_t b = 8; a = -((int64_t )(b * 2u)); the unsigned int result of (b * 2u) is converted to int64_t and that is negated. Andrew.
