Re: [PATCH RFC] rcu: change return type to bool

Nicholas Mc Guire
2015-05-24T08:46:56Z

On Sun, 24 May 2015, Joe Perches wrote:

> On Sun, 2015-05-24 at 10:10 +0200, Nicholas Mc Guire wrote:
> > On Sun, 24 May 2015, Joe Perches wrote:
> >
> > > On Sun, 2015-05-24 at 09:27 +0200, Nicholas Mc Guire wrote:
> > > > On Sat, 23 May 2015, Steven Rostedt wrote:
> > > []
> > > > > > - return sum;
> > > > > > + return !!sum;
> > > > >
> > > > > Hmm I wonder if gcc is smart enough to do the above without the need
> > > > > for !!? That is, will it turn to !! because the return of the function
> > > > > is bool, or does gcc complain about it not being bool without the !!?
> > > > > Not a criticism of the patch, just a curiosity.
> > > > >
> > > > gcc will not complain if you assign a unsigned long to a boolean
> > > > as I understand it it is a macro and is not doing any type
> > > > checking/promotion at all - so anything can be assigned to a bool
> > > > without warning (including double and pointers).
> > > > The !! will though always make the type compatible with int so it is
> > > > a well defined type atleast as far as __builtin_types_compatible_p()
> > > > goes, and !! also makes static code checkers happy (that are maybe not
> > > > as smart as gcc) and it does make the intent of sum being treated
> > > > as boolean here clear.
> > >
> > > 6.3.1.2 Boolean type
> > >
> > > When any scalar value is converted to _Bool, the result is 0 if the
> > > value compares equal to 0; otherwise, the result is 1.
> > >
> > As I understand this applies to arithmetic operations so for
> > bool x = false; int i = 42; x += i; x is defined to be true
> > but here it is the return type and not an arithmetic operation
> > so does this apply here without the !!?
>
> Yes, it does. return is an implicit conversion.
>
> 6.8.6.4 The return statement
>
> 3 If a return statement with an expression is executed, the value of
> the expression is returned to the caller as the value of the function
> call expression. If the expression has a type different from the
> return type of the function in which it appears, the value is
> converted as if by assignment to an object having the return type of
> the function.
>
get it - thanks for the clarification !

thx!
hofrat