具有不同功能长度的函数的传递参考

Question

I have a function with optional number of arguments, something like this:

function DoSomething()
{
    $args = funct_get_args();

    // and the rest of function
}

In the function above, how can I define the first argument to be passed by reference?

So when I calling it, I be able to do so:

DoSomething(&$first, $second, $third);

Answer 1

Simply declare it in the parameter list:

function DoSomething(&$first)
{
    $args = func_get_args();

    // and the rest of function
}

DoSomething($first, $second, $third);