On 6 February 2018 at 15:50, Peter T. Breuer <ptb@xxxxxxxxxxxxxx> wrote: > "Also sprach Liu Hao:" >> Well you could have declared the `printf()` function: > > Admitted, m'lud. But that would have meant an extra line of code, and I > am miserly. > >> extern int printf(const char *restrict format, ...); >> >> > % ./a.out >> > 0xe0000001 >> > ^ SIGNED right shift >> > >> > OK, so x>>y was done signed. That means both x,y were converted to >> > signed int (trivially). >> >> No. See below. > > I do not agree with the point made below, ingenious though it seems to > me, hinging (apparently) on the meaning of "the". But let us see ... > >> > First: NO INTEGER PROMOTION is done, by this rule >> > >> > any operand whose type ranks lower than int is automatically converted >> > to the type int, provided int is capable of representing all values of >> > the operand's original type. If int is not sufficient, the operand is >> > converted to unsigned int. >> > >> > Both x and y are int, so do not rank "lower than int". Rule does not >> > apply. All as is. So INTEGER CONVERSION applies from there: >> >> It doesn't. >> >> I don't have C99 at hand, so I will quote C11 (WG14 N1570) for you: >> >> ``` >> 6.5.7 Bitwise shift operators >> ... >> Semantics >> 3 The integer promotions are performed on each of the operands. ... >> ``` >> >> Note that the standard references _the integer promotions_, in contrast >> to what you see elsewhere: > > I am afraid that "the" may refer to something that does not exist > but is properly identified, such as "the large white rabbit that > I met yesterday between the third and fourth pubs". It's a definite > article, not a claim of cardinality at least one. > > That interpretation is helped by the "S" on the end of "promotionS", > because in the case of an empty set of things, we still refer to the > things in the empty set as being thingS [whereas I grant you it > would be odd to refer to the thing (singular) in the empty set]. > > In other words, the plural allows for 0,1,2,... number of promotions > to be applied. I can find no rule that applies a promotion, so I > contend the answer is 0. > > The horses that almost ran me over yesterday are a figment of my > imagination. > >> 6.5.5 Multiplicative operators >> ... >> Semantics >> 3 The usual arithmetic conversions are performed on the operands. > > Those are conversions, not promotions. I aver that zero PROMOTIONS > and one CONVERSION (to UNSIGNED int) should be applied by the > rule I quoted. > >> ``` >> 6.5.9 Equality operators >> ... >> Semantics >> 4 If both of the operands have arithmetic type, the usual arithmetic >> conversions are performed. > > Ditto. > >> Clearly, the type of the operand on the right-hand side is not altered - >> it remains `unsigned int`. > > Yes. And arithmetic operations are applied to operands of the same > type. That's what conversion is for: > > The usual arithmetic conversions are rules that provide a mechanism > to yield a common type when both operands of a binary operator are > ^^^^^^^^^^^^^^ > balanced to a common type ... > > Ergo, one was unsigned, so they should both have been. No, because the operands of bitshift operators aren't balanced to a common type. You asked for clarification and have got your answer, but seem determined to stick to your original interpretation.
