Re: Top 10 most common hard skills listed on resumes...

On 31/08/2024 23:31, Keith Thompson wrote:

Bart <bc@freeuk.com> writes:
[...]

Given this:
>
    x + y = z;
    (x + y) = z;
>
My compiler produces the same AST for the LHS of '=' in both cases.

[...]
 If that's the case (and I don't doubt that it is), then your compiler is
not following the grammar specified by the ISO C standard.  Since
`x + y` is not a unary-expression, `x + y = z` is not a syntactically
valid assignment expression.

Yet no compiler out of the dozen I tried reported a syntax error (except SDCC).
So what AST is produced by gcc, if any?
Most compilers, including mine, complain that an lvalue is expected.

 A parser that strictly follows the ISO C grammar would reject
(diagnose, flag, whatever) `x + y = z;` just as it would reject `x = y +;`.

Which one does that (apart from SDCC)?

This is an observation, not a complaint.  It doesn't imply that your
compiler is non-conforming or buggy.  A parser that doesn't strictly
follow the ISO C grammar could still be part of a conforming compiler.
 

I can also say that the C grammar is buggy:
     assignment-expression:
         conditional-expression
         unary-expression asssignment-operator assignment-expression
When attempting to parse an assignment-expression, do you go for a conditional-expression or unary-expression?
The latter is a subset of the former. If you go for a conditional-expression and find that an assignment-operator follows, now you have to perform some analysis on the LHS to see if that conditional-expression contains only a unary-expression.
However, if it's not a unary-expression, it will fail for other reasons
anyway, because all those other things that a conditional-expression will be, can't be lvalues.
That also applies to many unary-expressions, such as 42, or a++; those can't be lvalues either, even though the syntax is valid.
So it makes sense to do only an lvalue test.