OR Preference changing with a return

问题

From what I see operator precedence makes sense in these two examples:

$a = false;
$b = true;
$c = $a || $b;

Here $c is true

$a = false;
$b = true;
$c = $a or $b;

Here $c is false

---

I understand the reasoning behind it. However the following:

$a = false;
$b = true;
return $a or $b;

Returns true, which puzzles me.

What is the reason for this?

回答1:

or has lower precedence than =, so this:

$c = $a or $b;

Becomes this:

($c = $a) or $b;

But this doesn't make sense:

(return $a) or $b;

So you get this:

return ($a or $b);

回答2:

Within an expression, operator precedence applies. =, || and or are all operators and $c = $a or $b is an expression. And according to operator precedence it evaluates as ($c = $a) or $b.

However, return is a statement. return is not an operator and does not group by operator precedence. It always evaluates as return <expression>, and therefore always as return ($a or $b).

The result of the expression $c = $a or $b is true BTW. $c is being assigned false in the course of the expression, but the expression overall returns the value true ($b). So even this would return true:

return $c = $a or $b;

来源：https://stackoverflow.com/questions/51520435/or-preference-changing-with-a-return