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>
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