CODEWITHSUNDEEP
phpanonymous-function
Faiz Mohamed Haneef
Faiz Mohamed Haneef

Reputation: 3596

PHP anonymous functions chaining

$app = 'App here';


$fn1 = function($var) use($app){
    $fn2($var);
};

$fn2 = function($var) use($app){
    echo $var;
};

$fn1('variable');

In the above example, I am trying to chain/forward multiple anonymous functions. However, at the below line I receive an error "Notice: Undefined variable: fn2"

$fn2($var)

How do I achieve the chaining of anonymous functions.

Upvotes: 2

Views: 606

Answers (4)

Sherif
Sherif

Reputation: 11942

Just to clarify a few things first, chaining is something traditionally attributed to the Object Oriented Paradigm (as in method chaining), whereby one method returns an instance in order to chain multiple method calls together on the return value of each successive method.

In a functional paradigm what you're doing is more commonly referred to as a higher order function - or a function whose argument takes another function or returns a function.

The problem with your existing code is that functions in PHP do not implicitly import variables from the global scope (see variable scoping in PHP) or any other block scope. The only caveat being that they do gain access to $this or the object instance when they are defined within object context (i.e. inside of a class method that is not static). At least as of PHP 5.4.

So instead you must explicitly import variables from the global or local scope outside of the closure by using the use statement, which you have used to import the variable $app in your example. Though, importing variables touching the outer scope may not always be an elegant solution within a functional paradigm, assuming that is what you're going for.

So in the essence of a higher order function you can more succinctly express your code in the following manner...

$app = 'App here';


$fn1 = function($fn2, $var) use($app){
    $fn2($var);
};

$fn2 = function($var) use($app){
    echo $var;
};

$fn1($fn2, 'variable');

Which would give you the output variable.

Upvotes: 3

TheTrueTDF
TheTrueTDF

Reputation: 184

You need to define $fn2 before you use the variable

EDIT: and to pass the variable in the use()

$app = 'App here';

$fn2 = function($var) use($app){
    echo $var;
};

$fn1 = function($var) use($app, $fn2){
    $fn2($var);
};

$fn1('variable');

Upvotes: 1

Rizier123
Rizier123

Reputation: 59701

The problem is, that your variable $fn2 is out of scope in your first anonymous function. So just put the assignment of $fn2 before $fn1 and then pass the variable as you already did in the use(), e.g. $fn1 = function($var) use($app, $fn2){

Upvotes: 3

gastonmancini
gastonmancini

Reputation: 1102

The problem is that you are not passing the $fn2 as a parameter in the use statement of the closure.

Try the following code:

    $app = 'App here';

    $fn2 = function($var) use($app){
       echo $var;
    };

    $fn1 =  function($var) use($app, $fn2){
       $fn2($var);
    };

    $fn1('variable');

Here you have your example working in an online php tester.

Upvotes: 5

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