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

[Paul E. McKenney]

On Sun, May 24, 2015 at 10:46:50AM +0200, Nicholas Mc Guire wrote:
> 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 !

Hello, Nicholas,

Were you planning to send an updated patch?

							Thanx, Paul

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