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

On 31/08/2024 23:10, Tim Rentsch wrote:

Bart <bc@freeuk.com> writes:
 

On 30/08/2024 21:41, Keith Thompson wrote:
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Tim Rentsch <tr.17687@z991.linuxsc.com> writes:
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Kaz Kylheku <643-408-1753@kylheku.com> writes:
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On 2024-08-29, Ben Bacarisse <ben@bsb.me.uk> wrote:
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Bart <bc@freeuk.com> writes:
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I think that these (with x, y having compatible scalar types):
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     x + 1 = y;
     (x + 1) = y;         // in case above was parsed differently
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are both valid syntax in C. It will fail for a different reason:
an '+' term is not a valid lvalue.

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The compiler must tell you that neither is valid C.  That's
because what is required on each side of assignment is not
exactly the same thing.  It's a distraction to argue about why
each is not valid C as both have errors that require diagnostic
at compile time.

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Bart is only saying that it's valid syntax, not that it's valid C.
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According to the ISO C syntax (not taking into account contraints,
which are not syntax) that view is justified.

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The second line is syntactically well-formed.  The first line is
not.

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Right, because the LHS of an assignment is a unary-expression.
`(x + 1)` can be parsed as a unary-expression, but `x + 1` cannot.

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

 To understand why they are different, try drawing parse trees
rather than abstract syntax trees.
 https://en.wikipedia.org/wiki/Parse_tree

Yeah, the CST differs in retaining the parentheses. But I can already see that from the source code.
For normal compilation, that information is redundant.