Call function within a function PHP

Question

1    function foo($i){
2       return bar($i)*4;
3       function bar($i){
4           return $i*4;
5           }
6       }
7    echo foo(4);

return

Fatal error: Call to undefined function bar() in /var/www/index.php on line 2

why doesn't it work? it works well in javascript, while it works when i do this:

function foo($i){
   return bar($i)*4;
   }

function bar($i){
   return $i*4;
   }

The first doesn't work in JS either. And second works in JS too. (But note that the first, in PHP, doesn't create a locally-scoped closure but simply a global function IIRC). — user395760
– user395760, Commented Jun 25, 2011 at 17:00
Why do you expect a thing working in one language to work in another one? — Felix Kling
– Felix Kling, Commented Jun 25, 2011 at 17:01
@delnan: It does work in JavaScript, because of hoisting. Function and variable declarations are hoisted to the top of the function. — Felix Kling
– Felix Kling, Commented Jun 25, 2011 at 17:02
that wouldn't work in any languages... code after return.... — dynamic
– dynamic, Commented Jun 25, 2011 at 17:02
@Felix: Ah, yes. I keep forgetting this ugly inconsistency between var f = function () {} and function f() {} (wishful thinking?). — user395760
– user395760, Commented Jun 25, 2011 at 17:04

Percy · Accepted Answer · 2011-06-25 17:03:17Z

21

Define the function above your return value, otherwise it never gets executed.

<?php
function foo($i){
    function bar($i){
        return $i*4;
    }
    return bar($i)*4;
}
echo foo(4);
?>

answered Jun 25, 2011 at 17:03

Percy

1,0578 silver badges7 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

elliotrock Over a year ago

Yes order of events, saw that immediately

vascowhite · Accepted Answer · 2011-06-25 17:01:07Z

3

It doesn't work as you are calling bar() before it has been created. See example 2 here:- http://www.php.net/manual/en/functions.user-defined.php

answered Jun 25, 2011 at 17:01

vascowhite

18.5k9 gold badges64 silver badges79 bronze badges

Comments

maaudet · Accepted Answer · 2011-06-25 17:01:00Z

1

Your code never reach function bar(...), therefore it's never defined.

You need to put your bar() function before your foo() or before the return bar. (Answer based on your first example).

With your second example you probably define bar() before you use foo() therefore bar() is defined and it works fine.

So as long as you have your function defined when you hit a certain spot in your code it works fine, no matter if it's called from within another function.

answered Jun 25, 2011 at 17:01

maaudet

2,3584 gold badges20 silver badges29 bronze badges

Comments

Tarek · Accepted Answer · 2011-06-25 17:01:24Z

1

If you define function within another function, it can be accessed directly, but after calling parent function. For example:

function a () {
    function b() {
    echo "I am b.";
   }
echo "I am a.<br/>";
}
//b(); Fatal error: Call to undefined function b() in E:\..\func.php on line 8
a(); // Print I am a.
b(); // Print I am b.

answered Jun 25, 2011 at 17:01

Tarek

3,7883 gold badges42 silver badges62 bronze badges

Comments

Rajasekar Gunasekaran · Accepted Answer · 2011-06-25 17:08:07Z

0

Execution not execute after return statement

change your code like this

 function foo($i){
 function bar($i){
  return $i*4;
 }
 return bar($i)*4;     
}
echo foo(4);

edited Jun 25, 2011 at 17:08

answered Jun 25, 2011 at 17:02

Rajasekar Gunasekaran

1,8393 gold badges27 silver badges43 bronze badges

Comments

jpgerb · Accepted Answer · 2022-04-29 22:40:08Z

0

For the sake of having a "recent" answer, here's what I did:

function printRx($conn, $patient)
{
    function rows($rows, $tod)
    {
        for ($x = 0; $x < count($tod); $x++) {
            if ($tod[$x] == 1) {
                $rows++;
                break;
            }
        }
        return $rows;
    }
    $tod['prn'] = (explode('|', $row['prn'])) ?? null;
    $rows = rows($rows, $tod['prn']);
    return;
}

Works beautifully

answered Apr 29, 2022 at 22:40

jpgerb

1,1221 gold badge11 silver badges42 bronze badges

Collectives™ on Stack Overflow

Call function within a function PHP

6 Answers 6

1 Comment

Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

1 Comment

Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related