access to array outside function - php

Question

Have found 2 threads on this here at Stackoverflow, hovever these are about how to return an array. I can generate an array inside the function and it is having the right content, however when doing var_dumping of $target_file outside the function it is NULL. Otherwise the code is doing what it is supposed to do.
Is it a scope thing? or...have I done something completely wrong here?
Can anyone please help me access the returned array outside the function?
function copy_files3($requested, $src_path, $send_path){
//function copy_files renames and copies pdf-files to a specific folder.
//ARRAY $requested (keys INT), (names STR) holds the names of the selected files
//STR $src_path is the full path to the requested files
//STR $send_path is the full path to the re-named files
//ARRAY $target_file holds the names of the renamed files
$i=0;
$target_file = array();
$src_filename = array();
$b=array();
foreach($requested as $value) {
//$value holds the names of the selected files.
//1 Expand to get the full path to the source file
//2 Generate a 10 char + .pdf (aka 14 char) long new file name for the file.
//3 Generate full path to the new file.
$src_filename[$i] = $src_path.$value;
$rnam[$i] = randomstring(); //function randomstring returns a 10 char long random string
$target_file[$i] = $send_path.$rnam[$i].'.pdf';
echo 'target_file['.$i.'] = '.$target_file[$i].'<br>';
copy($src_filename[$i],$target_file[$i]);
$i++;
}
return($target_file);
}
I have files renamed and placed in the correct folders on my server, the only problem is accessing the $target_file array after this function.

Answer 1

$target_file exists only inside the function. And yes, it's a scope thing.
You are returning this variable's value, but not the variable itself.
All you have to do is to assign returned value to some variable while invoking the function.
$returnedValue = copy_files3($requested, $src_path, $send_path);
Now, $returnedValue has the same value as $target_file had inside the function.

Answer 2

<?php
function copy_files3($requested, $src_path, $send_path){
//function copy_files renames and copies pdf-files to a specific folder.
//ARRAY $requested (keys INT), (names STR) holds the names of the selected files
//STR $src_path is the full path to the requested files
//STR $send_path is the full path to the re-named files
//ARRAY $target_file holds the names of the renamed files
$i=0;
$target_file = array();
$src_filename = array();
$b=array();
foreach($requested as $value) {
//$value holds the names of the selected files.
//1 Expand to get the full path to the source file
//2 Generate a 10 char + .pdf (aka 14 char) long new file name for the file.
//3 Generate full path to the new file.
$src_filename[$i] = $src_path.$value;
$rnam[$i] = randomstring(); //function randomstring returns a 10 char long random string
$target_file[$i] = $send_path.$rnam[$i].'.pdf';
echo 'target_file['.$i.'] = '.$target_file[$i].'<br>';
copy($src_filename[$i],$target_file[$i]);
$i++;
}
return($target_file);
}
$returnedValue = copy_files3($requested, $src_path, $send_path);
?>