The principle of PHP inference variable output

Question

<?php
$b="c";
$c=7;
$i=&$c+$$b;
var_dump($i);
?>

Why my output is int(7)?Can you tell me the reason?I think $i value is 7+7=14 but the result let me fuzzy.

Is there a reason you put the `&` before the `$c`? `$i=$c+$$b;` would give you 14. Maybe take a look at http://stackoverflow.com/questions/3774130/php-what-does-a-in-front-of-a-variable-name-mean. — chris85, Jan 05 '17 at 06:16
but even that , the value of $i should be $c+$$b .actually the output is 7 not 14 — Pulsar, Jan 05 '17 at 06:23
https://eval.in/709799, works for me, also all other versions seem to have the same result, https://3v4l.org/v0bs0 — chris85, Jan 05 '17 at 06:25
@chris85 you forgot the & in that test, if you copy his code exactly it doesn't work there either — Sjors Ottjes, Jan 05 '17 at 06:26
@SjorsOttjes No i didn't I told the OP to remove that. See comment 1 and/or comment 3. — chris85, Jan 05 '17 at 06:26
try var dumping this: &$c it will trow an error. That's why it isn't being added to the result — Lucas Meine, Jan 05 '17 at 06:31
@LukasMeine ues echo to instead of vardump its value is also 7 — Pulsar, Jan 05 '17 at 06:41

anwerj · Answer 1 · 2017-01-05T07:00:51.390

0

It is because of reference. Precedence Reference + string = Reference. Consider this.

$b = 7;
$a = &$b; // $a now points to $b;
$a = '?'; // The value where $a points is '?'
echo $b;  // 7

Now $a is pointer, a reference to $b; You can not add string on that.

The precedence of & is being preferred over +.
Check this for more detail

EDIT
One interesting test for better understanding.

$a = 7;
$d = ($c = &$a + 1); // It is composite expression;
echo $c; // again 7 because & is given preference to + 
echo $d; // 8 because that is the output of internal expression.

To be very frank, this is weirdly new to me.

edited Jan 05 '17 at 07:00

answered Jan 05 '17 at 06:33

anwerj

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but I output &$c its value is 7 $$b value is 7 so I think $i value will be 14.I can't understand your answer could you make it clearly?? – Pulsar Jan 05 '17 at 06:45

Déjà vu · Answer 2 · 2017-01-05T08:01:06.577

Replace

$i=&$c+$$b;

with

echo &$c+$$b;

and you get a syntax error. Doing

$d=3;
$i=&$c+$d;

you get also 7, but doing $i = $d+&$c; yields a syntax error. In your example, doing, then, $i++; increments both $i and $c.

The PHP parser expects VAR = REFERENCE; and stops after it gets the reference (i.e. it ignores $$b). Anyway doing REF + integer has a meaning in C derived languages, but not in PHP.

Note that & is not only the 'reference' operator, it's also the bitwise (AND) operator. The unary version is for references, the binary one is for bitwise. The language grammar is more complex for &, that could explain the problem.

That's a PHP bug. The parser should also yield an error in the example you give. Don't use that anyway as it has no meaning and will be likely / hopefully rejected in future releases.

The principle of PHP inference variable output

2 Answers2