CODEWITHSUNDEEP
phptype-hinting
brainimus
brainimus

Reputation: 11026

Error when passing string into method with type hinting

In the code below I call a function (it happens to be a constructor) in which I have type hinting. When I run the code I get the following error:

Catchable fatal error: Argument 1 passed to Question::__construct() must be an instance of string, string given, called in run.php on line 3 and defined in question.php on line 15

From what I can tell the error is telling me that the function is expecting a string but a string was passed. Why isn't it accepting the passed string?

run.php:

<?php
require 'question.php';
$question = new Question("An Answer");
?>

question.php:

<?php
class Question
{
   /**
    * The answer to the question.
    * @access private
    * @var string
    */
   private $theAnswer;

   /**
    * Creates a new question with the specified answer.
    * @param string $anAnswer the answer to the question
    */
   function __construct(string $anAnswer)
   {
      $this->theAnswer = $anAnswer;
   }
}
?>

Upvotes: 21

Views: 7724

Answers (5)

Axel
Axel

Reputation: 3323

NOTE

"type declarations" (aka "type hinting") are available for the following types since PHP 7.0.0:

  • bool The parameter must be a boolean value.
  • float The parameter must be a floating point number.
  • int The parameter must be an integer.
  • string The parameter must be a string.
  • bool The parameter must be a boolean value.

for the following types since PHP 7.1.0:

  • iterable The parameter must be either an array or an instanceof Traversable.

So from now on another answer to this question actually is (kind of):

Switch the PHP version to PHP7.x and the code will work as you expect.

http://php.net/manual/en/functions.arguments.php#functions.arguments.type-declaration

Upvotes: 1

Daniel Egeberg
Daniel Egeberg

Reputation: 8382

PHP doesn't support type hinting for scalar values. Currently, it's only possible for classes, interfaces and arrays. In your case, it's expecting an object that is an instance of a "string" class.

There is currently an implementation supporting this in the SVN trunk version of PHP, but it's undecided if that implementation will be the one that gets released in future versions of PHP, or if it will be supported at all.

Upvotes: 28

Iacopo
Iacopo

Reputation: 4292

From the PHP documentation (http://php.net/manual/en/language.oop5.typehinting.php)

Type Hints can only be of the object and array (since PHP 5.1) type. Traditional type hinting with int and string isn't supported.

No way to hint strings, ints or any other primitive type

Upvotes: 2

Mark Baker
Mark Baker

Reputation: 212402

Type hinting can only be used for object data types (or arrays since 5.1), not for the basic types like string, integer, float, boolean

Upvotes: 4

Sarfraz
Sarfraz

Reputation: 382616

Just remove string from constructor (not supported) , it should work fine eg:

function __construct($anAnswer)
{
   $this->theAnswer = $anAnswer;
}

Working Example:

class Question
{
   /**
    * The answer to the question.
    * @access private
    * @var string
    */
   public $theAnswer;

   /**
    * Creates a new question with the specified answer.
    * @param string $anAnswer the answer to the question
    */
   function __construct($anAnswer)
   {
      $this->theAnswer = $anAnswer;
   }
}

$question = new Question("An Answer");
echo $question->theAnswer;

Upvotes: 8

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