On Tue, 29 Jul 2008, Linus Torvalds wrote:
quoted text
>
>   [...]                           since statement expressions are gcc 
> extensions, and as such the gcc people could make up any semantics they 
> want to them, including just defining that a statement expression with 
> an lvalue value is the same lvalue rather than any temporary). 

In fact, that does seem what gcc-4.x does. The way to tell is to do

	const int *x;

	({ *x }) = 1;

and it's (a) legal (assignments to non-lvalues wouldn't work) and (b) 
gives a nice warning about assignment to read-only location, which in turn 
implies that the compiler properly just peeled off the de-reference even 
though it was inside the statement expression.

IOW, at least in gcc-4.3 (and apparently in earlier gcc-4 versions, but 
not in gcc-3.4.5), a statement expression with an lvalue return value _is_ 
actually an lvalue.

But that also means that there is no difference what-so-ever between (x) 
and ({ x; }) in gcc-4. And in gcc-3 there is, because apparently in gcc-3 
a statement expression is never an lvalue (which is actually the sane 
thing, imho).

			Linus
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