use ++a as return, why this happen?

Question

recently I did a test on return reference function follow the official example here , but confused with the result when return '++$a' instead, it seems they should be same from the answer What's the difference between ++$i and $i++ in PHP? It seems related to the version of PHP.

php version

PHP 5.6.28 (cli) (built: Dec  6 2016 12:38:54)
Copyright (c) 1997-2016 The PHP Group
Zend Engine v2.6.0, Copyright (c) 1998-2016 Zend Technologies

code

function &fun(){
        static $var = 1;
        return ++$var;
}

function &bar() {
    static $var = 1;
    ++$var;
    return $var;
}

$var2 =& fun();
$var3 =& bar();
fun();
bar();

echo 'var2:', $var2; // 2  why is it 2 instead of 3?
echo 'var3:', $var3; // 3

@Fky You linked to the wrong "static". You want static variable scope — GordonM, Mar 20 '17 at 15:44
http://php.net/manual/en/language.variables.scope.php#language.variables.scope.static — Condorcho, Mar 20 '17 at 15:45
@Condorcho yeah, I have seen this, I am just confused with `return ++$var // why different result? ++$var; return $var` — Richard Li, Mar 21 '17 at 02:08

score 0 · Answer 1 · edited May 06 '18 at 19:49

0

You are using a reference to the same variable. So when you run bar() your $var is already 2.

edited May 06 '18 at 19:49

Pradeep

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answered Mar 20 '17 at 15:38

Sergei Kutanov

825
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The scope of $var is inside function, it is ok to replace $var with any other names – Richard Li Mar 21 '17 at 02:15

score -1 · Answer 2 · answered Mar 21 '17 at 02:44

-1

function &fun(){
        static $var = 1;
        return ++$var;
}

this function makes very little sense, because its return expression contains a value as the operand. The value is being returned after evaluating the unary prefix operator ++.

So, ++$var increments a variable, then returns a value of the variable, not a reference to a variable itself.

In short: references can be only obtained for a variables, taking a reference from a value does not make much sense.

answered Mar 21 '17 at 02:44

zerkms

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from php manual [Incrementing/Decrementing Operators](http://php.net/manual/en/language.operators.increment.php), `++$a, Increments $a by one, then returns $a.` why do you say that returns a _value_ of the variable. – Richard Li Mar 21 '17 at 03:19
@RichardLi because in php expressions returns values. File a bug so that they clarified that in the documentation, it clearly must be improved to avoid misunderstanding. – zerkms Mar 21 '17 at 08:22

use ++a as return, why this happen?

php version

code

2 Answers2