why then the code doesn't work?
the error I'm getting is: "Fatal error: Call to undefined function
solution1() in ..." even the function itself is just above the line that
calls function?

-afan



Stut wrote:
> Jay Blanchard wrote:
>> I don't think you can put a function name in a variable and call it like
>> $function($var).
>
> Yes you can - it's basically the same as variable variables.
>
> -Stut
>