Unexpected behaviour of current() in a foreach loop [duplicate]

Answer 1

FROM PHP.net

The current() function simply returns the value of the array element that's currently being pointed to by the internal pointer. It does not move the pointer in any way

then: use next()

$list = array("A", "B", "C","D");
foreach ($list as $var) {
    print(current($list));
    next($list);
}

NOTE: the first element will not be printed because foreach moved the pointer to second element of the array :)

This example will explain the full behaviour:

$list = array("A", "B", "C","D");
foreach ($list as $var) {
   if(!isset($a)) reset($list); $a = 'isset';
   print(current($list));
   next($list);
}

out put is ABCD

Please also note that:

As foreach relies on the internal array pointer changing it within the loop 
may lead to unexpected behavior.

foreach

---

EDIT: I want to share also my new confusing finding!!!

Example1:

$list = array("A", "B", "C","D");
$list_copy = $list;
foreach ($list as $key => $val) {
  current($list_copy);
  echo current($list);
  //next($list);
}

OUTPUT: AAAA

Example2:

$list = array("A", "B", "C","D");
$list_copy = $list;
foreach ($list as $key => $val) {
  current($list_copy);
  echo current($list);
  next($list);
}

OUTPUT: ABCD

When calling current() function inside foreach even for another array it will affect the foreach behavior...

Example3:

$list = array("A", "B", "C","D");

$refcopy = &$list;

foreach ($list as $key => $val) {
  if(!isset($a)) { $a = 'isset'; reset($list); }
  echo current($list);
  next($list);
}

OUTPUT: ACD (WOW! B is missing)

Example: 4

$list = array("A", "B", "C","D");

$refcopy = &$list;

foreach ($list as $key => $val) {
  echo current($list);
  next($list);
}

OUTPUT: BCD

It can't be decided exactly what will happen inside foreach loop!!!