> b) A function which returned a variable by reference couldn't be returned > by reference. I think this one is probably OK to code.
Do you mean a referenced variable returned by a function couldn't be returned as a reference by its caller? That is to say,
function &foo() { return $a; }
function &bar() { return foo(); }
$a = &bar();
Yeah there was a bug with this. It should work now.
Andi
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