language/references.xml
6d29533483657c036e49edb5ea88c7103d126681
6d29533483657c036e49edb5ea88c7103d126681
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@@ -74,7 +74,7 @@ $b = array();
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foo($b['b']);
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var_dump(array_key_exists('b', $b)); // bool(true)
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$c = new StdClass;
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$c = new stdClass;
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foo($c->d);
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var_dump(property_exists($c, 'd')); // bool(true)
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?>
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</note>
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<para>
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The same syntax can be used with functions that return
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references, and with the <literal>new</literal> operator (since PHP
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4.0.4 and before PHP 5.0.0):
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references:
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<informalexample>
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<programlisting role="php">
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<![CDATA[
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]]>
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</programlisting>
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</informalexample>
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Since PHP 5, <link linkend="language.oop5.basic.new">new</link>
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returns a reference automatically, so
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using <literal>=&</literal> in this context is deprecated and
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produces an <constant>E_DEPRECATED</constant> message in PHP 5.3 and
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later, and an <constant>E_STRICT</constant> message in earlier versions.
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(Technically, the difference is that, in PHP 5, object variables, much like
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resources, are a mere pointer to the actual object data, so these object
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references are not "references" in the same sense used before (aliases).
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For more information, see <link linkend="language.oop5.references">Objects
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and references</link>.)
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</para>
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<para>
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Using the same syntax with a function that does <emphasis>not</emphasis>
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return by reference will give an error, as will using it with the result
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of the <link linkend="language.oop5.basic.new">new</link> operator.
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Although objects are passed around as pointers, these are not the same as references,
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as explained under <link linkend="language.oop5.references">Objects and references</link>.
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</para>
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<warning>
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<para>
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<simpara>
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There is no reference sign on a function call - only on
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function definitions. Function definitions alone are enough to
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correctly pass the argument by reference. As of PHP 5.3.0,
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you will get a warning saying that "call-time pass-by-reference" is
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deprecated when you use & in <literal>foo(&$a);</literal>.
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And as of PHP 5.4.0, call-time pass-by-reference was removed, so
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using it will raise a fatal error.
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correctly pass the argument by reference.
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</simpara>
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</note>
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</para>
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</simpara>
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</listitem>
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<listitem>
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<simpara>
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New statements, i.e. <literal>foo(new foobar())</literal>
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</simpara>
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</listitem>
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<listitem>
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<para>
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References returned from functions, i.e.:
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<informalexample>
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$a = 5;
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return $a;
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}
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foo(bar()); // Produces fatal error since PHP 5.0.5
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foo(bar()); // Produces a notice
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foo($a = 5); // Expression, not variable
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foo(5); // Produces fatal error
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class Foobar
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{
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}
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foo(new Foobar()) // Produces a notice as of PHP 7.0.7
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// Notice: Only variables should be passed by reference
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?>
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]]>
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</programlisting>
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</informalexample>
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These requirements are for PHP 4.0.4 and later.
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</para>
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</sect1>
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work as you are attempting to return the result of an
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<emphasis>expression</emphasis>, and not a variable, by reference. You can
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only return variables by reference from a function - nothing else.
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Since PHP 4.4.0 in the PHP 4 branch, and PHP 5.1.0 in the PHP 5 branch, an
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<constant>E_NOTICE</constant> error is issued if the code tries to return
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a dynamic expression or a result of the <literal>new</literal> operator.
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</simpara>
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</note>
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<para>
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To use the returned reference, you must use reference assigment:
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To use the returned reference, you must use reference assignment:
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<informalexample>
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<programlisting role="php">
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<![CDATA[
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won't unset the global variable.
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</simpara>
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</sect2>
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<sect2 xml:id="references.this">
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<title><literal>$this</literal></title>
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<simpara>
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In an object method, <varname>$this</varname> is always a reference
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to the caller object.
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</simpara>
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</sect2>
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</sect1>
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</chapter>
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