On Mon, 2010-01-04 at 11:46 -0500, Greg Freemyer wrote:
> >
> > On Sat, Jan 2, 2010 at 8:10 AM, Shawn <citypw@xxxxxxxxx> wrote:
> >>
> >> hello guys,
> >> I got a newbie confused when I was looking into the source code of
> >> s3c2440's RTC driver.I dont know what is __v excatly means.anyone can
> >> tell?thanks anyway!
> >>
> >> #define readb(c) ({ __u8 __v = __raw_readb(__mem_pci(c)); __v; })
>
> On Sun, Jan 3, 2010 at 11:49 AM, Anand Arumugam <anand.arumug@xxxxxxxxx> wrote:
> > I think the last __v; inside the macro is to avoid compiler warning or error
> > that the unsigned 8-bit variable __v is not being used inside the scope
> > defined by the macro.
>
> No, I believe it is the value assigned to the statement sequence
> wrapped in the {} pair.
>
> ie. {statement1; statement2; variable} evaluates to the value of the
> variable which is the last statement.
No.
"{statement1; statement2; variable}" is pure (from K&R times ) C (if you
fix the syntax error of the missing ";" before the closing "}") and this
is usually called a "compound statement".
It becomes an expression if you
- use a gcc (young enough to implement it) and
- put the "( and ")" around it.
And the the result of the last statement becomes the result of the
expression
Bernd
--
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