wdseelig

Thank you guru!!

Or does it mean that functioncall is being called with two variables, the first of type array() and the second with value $variable?

of course I've seen default parameters, but that is not what the call to the function looks like [there is no = sign]. Does the (array(),$variable) construct mean that the function is being called with an array with a default value of $variable?

I am looking at the following code snippet: functioncall(array(),$variable) I've looked everywhere, but cannot find what this is supposed to mean. In the function that is called I see function functioncall($var = array(), $var1 = array()) Is this some kind of cast?? Thanks!