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;
 }

• user395760: 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).
• Felix Kling: Why do you expect a thing working in one language to work in another one?
• Felix Kling: @delnan: It does work in JavaScript, because of hoisting. Function and variable declarations are hoisted to the top of the function.
• dynamic: that wouldn't work in any languages... code after return....
• user395760: @Felix: Ah, yes. I keep forgetting this ugly inconsistency between var f = function () {} and function f() {} (wishful thinking?).

Answer 1

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);
?>

• elliotrock: Yes order of events, saw that immediately

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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);

Answer 6

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