Re: [sdcc-devel] Two questions about the yacc script sdcc.y

[Maarten B. <sou...@ds...>]

Erik,

> Both C99 and C11 (and probably the earlier specs as well, but I don't have
> them handy to check) specifically say the "expression ++E is equivalent
> to (E+=1)". Later when discussing the semantics of assignment operators,
> it says an "assignment expression has the value of the left operand after
> the assignment, but is not an lvalue". So I think SDCC should really be
> rejecting "++a = b" rather than accepting it as it currently does.

Thank you. I had read this part but still overlooked it :(
So, ++a is not an lvalue and should generate an error.

>> The order is not guaranteed? Are you saying C can interpret ++a=b as
>> a=b;++a ? Or even c=a+1;a=b;a=c ?
>>
>> Maarten
>
> The ordering has some flexibility, but there still some constraints. The C
> standard defines certain points in time in the execution of the program
> called "sequence points". Between consecutive sequence points, when there
> are multiple side effects (such as modifying a variable's value), these
> side effects are allowed to occur in any order.
>
> Without getting into the all of details of how sequence points are
> defined, a basic expression statement that does not call a function or use
> the logical AND, logical OR, conditional ( ? : ), or comma operators will
> be bounded by the same pair of sequence points and so the order in which
> any variables are modified is left to the compiler implementation. Some
> example expressions with ambiguous results:
>
>    x = (x = 1) + 1;          /* final value for x may be either 1 or 2 */
>
>    x = x++;  /* final value for x may or may not reflect the increment */
>
> The comma operator (as well as the other operators I noted above) does
> introduce a new sequence point, so any side effects in the left expression
> will complete (in some order) before any side effects in the right
> expression.
>
>    a = a + 1, a = b;  /* final value for a is always b */
>
> but as I mentioned, ++a is expanded as (a+=1) not (a=a+1,a)
>
>    Erik

So if ++a was an lvalue, it still contains no sequence point. And thus
++a=b could be interpreted as (?)
a=a+1, a=b
c=a+1, a=b, a=c
a=b, a=a+1

Maarten