"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.
(I'm willing to entertain ingenious interpretations of the word "type"
or "common" if that is what you are planning next).
Regards
PTB
