What does double question mark (??) operator mean in PHP

It's the "null coalescing operator", added in php 7.0. The definition of how it works is:

It returns its first operand if it exists and is not NULL; otherwise it returns its second operand.

So it's actually just isset() in a handy operator.

Those two are equivalent¹:

$foo = $bar ?? 'something';
$foo = isset($bar) ? $bar : 'something';

Documentation: http://php.net/manual/en/language.operators.comparison.php#language.operators.comparison.coalesce

In the list of new PHP7 features: http://php.net/manual/en/migration70.new-features.php#migration70.new-features.null-coalesce-op

And original RFC https://wiki.php.net/rfc/isset_ternary

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EDIT: As this answer gets a lot of views, little clarification:

¹There is a difference: In case of ??, the first expression is evaluated only once, as opposed to ? :, where the expression is first evaluated in the condition section, then the second time in the "answer" section.

$myVar = $someVar ?? 42;

Is equivalent to :

$myVar = isset($someVar) ? $someVar : 42;

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For constants, the behaviour is the same when using a constant that already exists :

define("FOO", "bar");
define("BAR", null);

$MyVar = FOO ?? "42";
$MyVar2 = BAR ?? "42";

echo $MyVar . PHP_EOL;  // bar
echo $MyVar2 . PHP_EOL; // 42

However, for constants that don't exist, this is different :

$MyVar3 = IDONTEXIST ?? "42"; // Raises a warning
echo $MyVar3 . PHP_EOL;       // IDONTEXIST

Warning: Use of undefined constant IDONTEXIST - assumed 'IDONTEXIST' (this will throw an Error in a future version of PHP)

Php will convert the non-existing constant to a string.

You can use constant("ConstantName") that returns the value of the constant or null if the constant doesn't exist, but it will still raise a warning. You can prepended the function with the error control operator @ to ignore the warning message :

$myVar = @constant("IDONTEXIST") ?? "42"; // No warning displayed anymore
echo $myVar . PHP_EOL; // 42