PHP functions 101 question, really simple

Question

I'm kind of slow in the head so can someone tell me why this isn't working:

function foo() {
 $bar = 'hello world';
 return $bar;
}
foo();
echo $bar;

I just want to return a value from a function and do something with it.

• Maxpm: I don't know PHP, but have you tried echo foo();?

Answer 1

Because $bar does not exist outside of that function. Its scope is gone (function returned), so it is deallocated from memory.

You want echo foo();. This is because $bar is returned.

In your example, the $bar at the bottom lives in the same scope as foo(). $bar's value will be NULL (unless you have defined it above somewhere).

Answer 2

Your foo() function is returning, but you're not doing anything with it. Try this:

function foo() {
 $bar = 'hello world';
 return $bar;
}
echo foo();
// OR....
$bar = foo();
echo $bar;

Answer 3

You aren't storing the value returned by the foo function Store the return value of the function in a variable

So

foo() should be replaced by

var $t = foo();
echo $t;

Answer 4

As the others have said, you must set a variable equal to foo() to do stuff with what foo() has returned.

i.e. $bar = foo();

You can do it the way you have it up there by having the variable passed by reference, like so:

function squareFoo(&$num) //The & before the variable name means it will be passed by reference, and is only done in the function declaration, rather than in the call.
{
 $num *= $num;
}
$bar = 2;
echo $bar; //2
squareFoo($bar);
echo $bar; //4
squareFoo($bar);
echo $bar; //16

Passing by reference causes the function to use the original variable rather than creating a copy of it, and anything you do to the variable in the function will be done to the original variable.

Or, with your original example:

function foo(&$bar) 
{
 $bar = 'hello world';
}
$bar = 2;
echo $bar; //2
foo($bar);
echo $bar; //hello world