Is explode equivalent to implode when using the same parameter?

Question

Let's say I have the following function:

public function normalize($string) {
  $substrings = explode(",", $string);
  return implode(",", $substrings);
}

Will ($string == normalize($string)) always be true? is there any special case I should consider?

Welcome to Stack Overflow, you have to show a problem or an issue about your question. — SaidbakR, Jan 11 '18 at 21:28
`$string != normalize($string)` if `$string` can't be cast to a string. — Alex Howansky, Jan 11 '18 at 21:30
`$string != normalize($string)` if `$string` is an object that implements the `__toString()` magic method. — Alex Howansky, Jan 11 '18 at 21:38
`$string != normalize($string)` if `$string` is a float with a [tiny floating point error](https://3v4l.org/6dPGg). — Alex Howansky, Jan 11 '18 at 21:50
@AlexHowansky It is equal, tried it, works because for $object with __toStrring, $object == $object . '' — Máthé Endre-Botond, Jan 12 '18 at 00:16
@SinistraD Dang, you're right, I must have inadvertently had a `===` in my test code. — Alex Howansky, Jan 12 '18 at 03:01

Máthé Endre-Botond · Accepted Answer · 2018-01-12T00:16:37.340

5

If $string is a string, yes.

Otherwise type conversion may occur:

implode(",", explode(",", 0))

This will result in "0" thus $string !== normalize($string) but $string == normalize($string) still holds true.

edited Jan 12 '18 at 00:16

answered Jan 11 '18 at 21:35

1 Answers1