Why does negation happen last in an assignment expression in PHP?

Question

The negation operator has higher precedence than the assignment operator, why is it lower in an expression?

e.g.

if (!$var = getVar()) {

In the previous expression the assignment happens first, the negation later. Shouldn't the negation be first, then the assignment?

Answer 1

The left hand side of = has to be a variable. $var is a variable, whereas !$var is not (it's an expr_without_variable).

Thus PHP parses the expression in the only possible way, namely as !($var = getVar()). Precedence never comes to play here.

An example of where the the precedence of = is relevant is this:

$a = $b || $c // ==> $a = ($b || $c), because || has higher precedence than =
$a = $b or $c // ==> ($a = $b) or $c, because or has lower precedence than =

Answer 2

In short, assignments will always have precedence over their left part (as it would result in a parse error in the contrary case).

 <?php
 $b=12 + $a = 5 + 6;
 echo "$a $b\n";
 --> 11 23

 $b=(12 + $a) = (5 + 6);
 echo "$a $b\n";
 --> Parse error

The PHP documentation has now a note concerning this question: http://php.net/manual/en/language.operators.precedence.php (i guessed it was added after your question)

Although = has a lower precedence than most other operators, PHP will still allow expressions similar to the following: if (!$a = foo()), in which case the return value of foo() is put into $a