timer-of: don't use conditional expression with mixed 'void' types
Randy Dunlap reports on the sparse list that sparse warns about this
expression:
of_irq->percpu ? free_percpu_irq(of_irq->irq, clkevt) :
free_irq(of_irq->irq, clkevt);
and honestly, sparse is correct to warn. The return type of
free_percpu_irq() is 'void', while free_irq() returns a 'const void *'
that is the devname argument passed in to the request_irq().
You can't mix a void type with a non-void types in a conditional
expression according to the C standard. It so happens that gcc seems to
accept it - and the resulting type of the expression is void - but
there's really no reason for the kernel to have this kind of
non-standard expression with no real upside.
The natural way to write that expression is with an if-statement:
if (of_irq->percpu)
free_percpu_irq(of_irq->irq, clkevt);
else
free_irq(of_irq->irq, clkevt);
which is more legible anyway.
I'm not sure why that timer-of code seems to have this odd pattern. It
does the same at allocation time, but at least there the types match,
and it makes sense as an expression.
Reported-by: Randy Dunlap <rdunlap@infradead.org>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
This commit is contained in:
@@ -25,7 +25,9 @@ static __init void timer_of_irq_exit(struct of_timer_irq *of_irq)
|
||||
|
||||
struct clock_event_device *clkevt = &to->clkevt;
|
||||
|
||||
of_irq->percpu ? free_percpu_irq(of_irq->irq, clkevt) :
|
||||
if (of_irq->percpu)
|
||||
free_percpu_irq(of_irq->irq, clkevt);
|
||||
else
|
||||
free_irq(of_irq->irq, clkevt);
|
||||
}
|
||||
|
||||
|
||||
Reference in New Issue
Block a user