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

[Joe Perches]

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.

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