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Re: Operator precedence [message #185054 is a reply to message #185050] Tue, 25 February 2014 04:45 Go to previous messageGo to previous message
Ben Bacarisse is currently offline  Ben Bacarisse
Messages: 82
Registered: November 2013
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Thomas 'PointedEars' Lahn <PointedEars(at)web(dot)de> writes:

> Ben Bacarisse wrote:
>
>> Thomas 'PointedEars' Lahn <PointedEars(at)web(dot)de> writes:
>>> Ben Bacarisse wrote:
>>>> Thomas 'PointedEars' Lahn <PointedEars(at)web(dot)de> writes:
>>>> > Ben Bacarisse wrote:
>>>> >> […] Expressions are built from operators, but is ',' an operator in
>>>> >> [PHP?
>>>> >> It appears in the operator precedence table but it isn't an operator
>>>> >> in an sense that would normally be understood by someone familiar with
>>>> >> these terms.
>>>> >
>>>> > ECMAScript has a “Comma Operator”, too.
>>>>
>>>> But my point was the PHP doesn't have one --
>>>
>>> But it does:
>>>
>>> for ($i = 0, $j = 42; $i < $j; ++$i);
>>
>> But try this:
>>
>> for (($i = 0, $j = 42); $i < $j; ++$i);
>
> Bogus example.
>
>> In general, putting an expression in parentheses does not stop it being
>> an expression.

This is why it's not bogus. By all means explain why your understanding
of what an operator permits the construction of something that is not an
expression, but don't just say "bogus".

I contend that for ',' to be an operator, it should form part of an
expression, and I contend that expressions can be put in parentheses and
remain expressions. You can challenge either notion, but then I think
we will be disagreeing only about definitions.

<snip>
>>> Please define what you consider to be an operator in a “normal sense of
>>> the word”.
>>
>> It's a syntactic symbol that can be used to combine simpler expressions
>> in such a way that the result is also, syntactically, an expression.
>
> A very narrow definition of “operator”.

Maybe, but it comports with what PHP considers to be an operator:

"An operator is something that takes one or more values (or
expressions, in programming jargon) and yields another value (so that
the construction itself becomes an expression)."

If you have a different definition, we could just agree on the fact that
PHP has no ',' operator as PHP defines the term.

What is your definition?

--
Ben.
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