Re: ambigious ++ expression silently accepted

[Andrew J. Schorr]

On Tue, Dec 13, 2005 at 12:27:48AM -0800, Paul Eggert wrote:
> Sorry, I don't see how mere precedence resolves the ambiguity here.
> If it were only a matter of precedence, the Awk expression foo++bar
> could be parsed either as ((foo++) bar), or as (foo (++bar)).  In both
> parses, the ++ would be unary, and would have higher precedence than
> concatenation.
> 
> I looked through the grammar defined by POSIX in
> <http://www.opengroup.org/onlinepubs/000095399/utilities/awk.html>
> and it appears to me that foo++bar is indeed ambiguous, since it can
> be parsed in two different ways:
> 
> non_unary_expr non_unary_expr -> lvalue INCR non_unary_expr
> non_unary_expr non_unary_expr -> non_unary_expr INCR lvalue

But that URL contains a table entitled: 
"Table: Expressions in Decreasing Precedence in awk".  And
that table clearly lists "Pre-increment" as having higher
precedence than "Post-increment".  So shouldn't the
expression be parsed as (foo (++bar))?  Or am I
interpreting the table improperly?

Note that down below after the YACC grammar, there's a note
that says:

"This grammar has several ambiguities that shall be resolved as follows:
* Operator precedence and associativity shall be as described in Expressions in
Decreasing Precedence in awk."

And in fact, gawk seems to parse as ((foo++) bar).

So is this a bug in gawk?

   $ gawk -v foo=1 -v bar=2 'BEGIN {print foo++bar}'
   12

I would expect the output to be "13" based on
the precedence table.

I note that Solaris 8 nawk behaves the same way as gawk:

   $ nawk -v foo=1 -v bar=2 'BEGIN {print foo++bar}'
   12

Regards,
Andy