关于php中的逻辑运算符&&

Question

here is the code

$a = 2;
$a && $a = 1;

var_dump( $a ); // result 1

Why $a is 1?

According to the document, logical operator '&&' has greater precedence than assign operator, it should be interpreted as ($a && $a) = 1, then there should be a syntax error.

Answer 1

From the documentation

Note:
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 .

The documentation doesn't specify what rules are followed at this point, but we can infer that your code is equivalent to this:

$a && ($a = 1)

Because && is "lazy" in that it won't bother evaluating further arguments if it finds one that's false, this means that $a will only be set to 1 if it previously held a truthy value (in your case, 2 is truthy). If you had set $a = 0 then it would have stayed 0 .

Answer 2

The way it's written, you're saying the same thing as this

$a = 2;
if($a) $a = 1;

Since 2 is something , it succeeds and changes $a to 1.

So why does it do this? You have a statement that $a = 2 ; There's no order of operations here so PHP processes it. The order of operations DOES come into effect on the second statement. Remember, $a is SOMETHING, (ie truthy). Let's change your code slightly

$a = 0;
$a && $a = 1;
echo $a;

As you can see, $a is now falsy. So the order prevents the script from changing the value and the script outputs 0

Answer 3

$a && $a = 1;

is an expression of boolean type. Since $a is truthy, the evaluation of the expression continues onto the second part, which is the assignment. As a side effect of the assignment, $a now holds 1 .

If $a had been 0 in the start, it'd remain 0 in the end, as the evaluation of the expression'd be terminated right after testing the first condition.