php - How does a reference in a for-each loop change an element of an array? -

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• php pass reference error after using same var 3 answers

$arr=array('a'=>'first','b'=>'second','c'=>'third'); foreach($arr &$a); foreach($arr $a); print_r($arr); 

the above code changes last element of $arr ['c'=>'second']. how that?

you creating references in first loop. @ end of loop $a still reference last value in array:

$arr = ['a','b','c']; foreach($arr &$a); debug_zval_dump($arr,$a); 

---

array(3) refcount(2){   [0]=>   string(1) "a" refcount(1)   [1]=>   string(1) "b" refcount(1)   [2]=>   &string(1) "c" refcount(2) } 

so, assigning $a change value:

//... previous code, , then: $a = 'i still reference'; debug_zval_dump($arr); 

---

array(3) refcount(2){   [0]=>   string(1) "a" refcount(1)   [1]=>   string(1) "b" refcount(1)   [2]=>   &string(22) "i still reference" refcount(2) } 

... so, if second foreach, happen: last item in array take on value of first item, original value lost, , following items of array. however, when gets set (the last one), original value has been lost, , doesn't change anything, last value of array take on value of item before last:

foreach($arr $a){     debug_zval_dump($arr); } 

---

array(3) refcount(3){   [0]=>   string(1) "a" refcount(1)   [1]=>   string(1) "b" refcount(1)   [2]=>   &string(1) "a" refcount(2) } array(3) refcount(3){   [0]=>   string(1) "a" refcount(1)   [1]=>   string(1) "b" refcount(1)   [2]=>   &string(1) "b" refcount(2) } array(3) refcount(3){   [0]=>   string(1) "a" refcount(1)   [1]=>   string(1) "b" refcount(1)   [2]=>   &string(1) "b" refcount(2) } 

in short: when using references in loop unless have very reason not to:

foreach($array &$a){     // logic } unset($a); // removes reference, can't accidentally assign , thereby mutate $array itself.