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Subject: Re: [PHP-DEV] Re: power operator (again)
From: rowan.collins@gmail.com (Rowan Collins)

On 21/12/2013 10:29, Robert Stoll wrote:
> echo ( -3 * 2); // -6
> $x = -3;
> echo $x * 2; // -6 -> does not behave like (-3) * 2

This is an extremely poor choice of example: (-3) * 2 and -(3*2) both 
evaluate to -6 anyway, since unary minus is effectively multiplication 
by -1, and therefore commutative with multiplication.

However, the last line *does* behave like "(-3) * 2", because left 
operand of the * has already been calculated and stored in $x; it could 
have been calculated using "$x = 10 - 13", in which case you could see 
it as equivalent to "(10 - 13) * 2" - the brackets basically meaning 
"calculate this bit first before you even look at the rest of the 
expression".

> Nope, that's not true. 1+2 is just already evaluated and the result was stored in $x rather than the expression.

That is exactly what I was trying to say: the expression $x * 3 does not 
need any rules of operator precedence, because it has only one operator 
(unless you count the $ sigil as an operator); if $x was previously 
assigned based on some other expression, the details of that expression 
are entirely irrelevant, only their answer matters.

The point being that all this discussion of precedence only applies if 
the unary minus operator appears *in the same expression* as the power 
operator. But it's very unlikely that anyone would write "$sq = -3 ** 2" 
rather than just "$sq = 9"; more likely, they will write "$sq = 
$calculated_value ** 2", which has no ambiguity, and will work as 
expected for any value of "$calculated_value".

Regards,

-- 
Rowan Collins
[IMSoP]