how to determine from inside of function if variable passed to it was by reference or not?

function开发者_开发百科 do_sth_with_var($variable) {
  if (is_by_reference($variable)) {
    $variable = something($variable)
  }
  else {
    return something($variable);
  }
}

so for example if 'something' was strtoupper:

do_sth_with_var(&$str); // would make $str uppercase, but
$str = do_sth_with_var($str); // that way should it be done if ommitting that '&'

Disclaimer: This does not directly answer the OP's question, but I'm providing this answer in the hope that it will help him/her.

I'd say, provide a consistent interface. Declare the function parameter as reference:

function do_sth_with_var(&$variable) {

and it will always be a reference.

You could also to the contrary and always copy the value:

function do_sth_with_var($variable) {
    $val = $variable;
    // work with $val here
    return $val;
}

No built-in PHP function I know changes its behaviour based on whether you pass a reference or not. They are clearly defined to either treat an argument as reference or not and I would argue that users are used to this clear definitions and can deal with them.

For example, sort always sorts an array in place. You know that. So if you want to keep the original, you make a copy before the call.

It depends on your function whether it makes sense to perform an in-place operation or not. E.g. it makes sense for sort but I'd say it does not make sense for string processing.

Also, as you are using PHP 5.3, passing variables by reference at call-time is deprecated:

As of PHP 5.3.0, you will get a warning saying that "call-time pass-by-reference" is deprecated when you use & in foo(&$a);.

So you (or whoever) should not do that anyway.

calltime-pass-by-ref is considered deprecated as of 5.3.

apart from that ... answering the question academically; I don't think there is any way to determine that propperly, b/c there is no difference between a "reference" and a usual parameter. In PHP every variable is a pointer to section in the symbol table. A reference just makes another variable point to the same section.

in PHP there are no references as C knows them for example. every variable is a pointer and every "reference" too.

PS: @Gordon: I forgot about the Reflection-Classes. Of course, they work on a meta-level. My answer is more directed at how PHP actually deals with paramters and variables.

This topic is covered very well by the first comment here (official php manual)

to answer your question literally, you can try experimenting with debug_zval_dump. Refcounts will be different for by val and by reference parameters.

function xxx($a) {
    debug_zval_dump($a);
}

$b = 123;
xxx($b);   // long(123) refcount(4)
xxx(&$b);  // long(123) refcount(1)

In your specific case you don't need to know if the value is passed by reference or not you can do something like:

function do_sth_with_var($variable) {
  $variable = something($variable)
  return $variable;
}

It should do exactly what you ask in your question

I don't know any good and safe way to control that a passed parameter is a reference and not just a value inside the body of a function at running time.