Give a function the __LINE__ variable without needing to declare it

Question

So I am trying to make a variable class meaning I can dynamically name my variables. And I wanted to have a undeclared variable at line x on my code. I managed to make a function be called like so:

function ($varname, $line = 0) {
  // Example
  echo 'unknwon variable found at line '.$line;
}

But I wanted to be able to remove the line = 0 of the function and instead just give it the line without needing to call the function like:

exampleFunction('Name', __LINE__);

And instead call it like:

exampleFunction('Name');

And the __LINE__ variable be passed with it instead of needing to include it. - I tried making the function that was being called look like so:

exampleFunction($varname, __LINE__) {
  // Executed code.
}

Though that also didn't work out, chucking an error.

Answer 1

You can use debug_backtrace method in your exampleFunction to get the caller info with line.

http://php.net/manual/tr/function.debug-backtrace.php

Example (copied from php.net)

<?php
// /tmp/a.php dosyası

function a_test($str)
{
    echo "\nHi: $str";
    var_dump(debug_backtrace());
}

a_test('friend');
?>

<?php
// /tmp/b.php dosyası
include_once '/tmp/a.php';
?>

Which outputs:

Hi: friend
array(2) {
[0]=>
array(4) {
    ["file"] => string(10) "/tmp/a.php"
    ["line"] => int(10)
    ["function"] => string(6) "a_test"
    ["args"]=>
    array(1) {
      [0] => &string(6) "friend"
    }
}
[1]=>
array(4) {
    ["file"] => string(10) "/tmp/b.php"
    ["line"] => int(2)
    ["args"] =>
    array(1) {
      [0] => string(10) "/tmp/a.php"
    }
    ["function"] => string(12) "include_once"
  }
}