将字符串传递给具有类型提示的方法时出错 [英] Error when passing string into method with type hinting

问题描述

在下面的代码中，我调用了一个具有类型提示的函数(它恰好是构造函数).当我运行代码时，出现以下错误:

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:

可捕获的致命错误:传递给Question :: __ construct()的参数1必须是字符串的实例，该字符串是给定的，在第3行的run.php中调用并在问题中定义. php 在第 15

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

据我所知，错误告诉我该函数需要一个字符串，但是传递了一个字符串.为什么它不接受传递的字符串?

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;
   }
}
?>

推荐答案

只需从构造函数中删除string(不支持)，它应该可以正常运行，例如:

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

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

工作示例:

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;

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