Re: Writing a C Compiler: lvalues

lawrence.jones@siemens.com writes:
> Keith Thompson <kst-u@mib.org> wrote:
>>
>> So in (x + 5), x *isn't* an lvalue, even though it started out as one.
>
> I disagree -- x *is* an lvalue, but it's converted to the value stored in
> the object when the containing expression is evaluated.


A subtle distinction at best. As I wrote upthread, the standard says:


C99 6.3.2.1p2:


        Except when it is the operand of [list of operators deleted],
        an lvalue that does not have array type is converted to the value
        stored in the designated object (and is no longer an lvalue).


So in the above context, ``x'' *was* an lvalue, but "is no longer"
an lvalue.


The wording is a bit odd. When exactly does it stop being an lvalue?
Surely conversions happen during execution, conceptually at least,
(right?) but by that time lvalue-ness is no longer meaningful.


(Note to comp.compilers readers: this is *very* C-specific.)


--
Keith Thompson (The_Other_Keith) kst-u@mib.org <http://www.ghoti.net/~kst>

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