Function inside a function.?

问题:

This code produces the result as 56.

function x ($y) {
    function y ($z) {
        return ($z*2);
    }

    return($y+3);
}

$y = 4;
$y = x($y)*y($y);
echo $y;

Any idea what is going inside? I am confused.

回答1:

X returns (value +3), while Y returns (value*2)

Given a value of 4, this means (4+3) * (4*2) = 7 * 8 = 56.

Although functions are not limited in scope (which means that you can safely 'nest' function definitions), this particular example is prone to errors:

1) You can't call y() before calling x(), because function y() won't actually be defined until x() has executed once.

2) Calling x() twice will cause PHP to redeclare function y(), leading to a fatal error:

Fatal error: Cannot redeclare y()

The solution to both would be to split the code, so that both functions are declared independent of each other:

function x ($y) 
{
  return($y+3);
}

function y ($z)
{
  return ($z*2);
}

This is also a lot more readable.

回答2:

(4+3)*(4*2) == 56

Note that PHP doesn't really support "nested functions", as in defined only in the scope of the parent function. All functions are defined globally. See the docs.

回答3:

Not sure what the author of that code wanted to achieve. Definining a function inside another function does NOT mean that the inner function is only visible inside the outer function. After calling x() the first time, the y() function will be in global scope as well.

回答4:

This is useful concept for recursion without static properties , reference etc:

function getReqursiveItems($id){
    $allItems = array();

    function getItemms($parent_id){
       return DB::findAll()->where('`parent_id` = $parent_id');
    } 

    foreach(getItems($id) as $item){
         $allItems = array_merge($allItems, getItems($item->id) );
    }

    return $allItems;
}

回答5:

It is possible to define a function from inside another function. the inner function does not exist until the outer function gets executed.

echo function_exists("y") ? 'y is defined\n' : 'y is not defined \n';
$x=x(2);
echo function_exists("y") ? 'y is defined\n' : 'y is not defined \n';

Output

y is not defined

y is defined

Simple thing you can not call function y before executed x

回答6:

Your query is doing 7 * 8

x(4) = 4+3 = 7 and y(4) = 4*2 = 8

what happens is when function x is called it creates function y, it does not run it.