Ignore double assignment by reference if second is inside condition

[sirbrillig]

When there are two assignments by reference with use of the variable between them, we consider the first assignment to be unused since it is immediately replaced:

$foo = &$bar; // unused
$foo = &$baz;
echo $foo;

However, if the second assignment is inside a condition, then we can't know if it will be used so we must ignore the warnings there:

$foo = &$bar;
if ($something) {
  $foo = &$baz;
}
echo $foo;

This PR makes that change.

Fixes #231