From: Ian McCulloch (ianmcc_at_[hidden])
Date: 2005-10-14 14:49:47
Jim Hyslop wrote:
> Robert Ramey <ramey <at> rrsd.com> writes:
>> I suspected that the & operator
>> was chosen due to its operator precedence being at a particular spot in
>> the operator precedence hierarchy[...]
> Except that operator precedence only applies to "built-in" operators -
> user defined operators are function calls, and have that level of
Really? I thought the operator precedence was implemented by grammar rules.
How does the grammar know whether an operator is builtin or user defined?
I am pretty sure that a+b*c means a+(b*c) irrespective of whether a,b,c are
builtin or UDT.
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