Can I declare a variable without $ in PHP

Question

I saw this code in a PHP book (PHP architect, ZEND PHP 5 Certification guide page 141)

class foo{
  public $bar;
  protected $baz;
  private $bas;

  public var1="Test"; //String
  public var2=1.23; //Numericvalue
  public var3=array(1,2,3);
}

and it says

Properties are declared in PHP using one of the PPP operators, followed by their name:

Note that, like a normal variable, a class property can be initialized while it is being declared. However, the initialization is limited to assigning values (but not by evaluating expressions). You can’t,for example,initialize a variable by calling a function—that’s something you can only do within one of the class’ methods (typically, the constructor).

I can not understand how var1, var2, var3 are declared. Isn't it illegal?

Answer 1

The sample code is (almost) valid (it's just missing a few $ signs.)

class foo
{
    // these will default to null
    public $bar;
    protected $baz;
    private $bas;

    // perfectly valid initializer to "string" value
    public $var1 = "Test"; //String

    // perfectly valid initializer to "float" value
    public $var2 = 1.23;    //Numericvalue

    // perfectly valid initializer to "array" value
    // (array() is a language construct/literal, not a function)
    public $var3 = array(1,2,3);
}

So, the book your code comes from is definitely in error.

Answer 2

No, this is an error. Defining:

public var1="Test"; //String

Will give you:

Parse error: syntax error, unexpected T_STRING, expecting T_VARIABLE

For details, see http://codepad.org/meMrSmfA.

Answer 3

Variables in PHP are "represented by a dollar sign followed by the name of the variable". Although dollarless variables have been requested, I doubt whether they we ever see them enabled.

Point in short: your code is invalid.

Answer 4

In php variable are auto casting.whatever you want keep in a variable no need to declare it type . But one mandatory thing is that when you going to declare a Variable in php you must have to use "$" which one you missed .Book declaration is

  public var1="Test"; //String
  public var2=1.23; //Numericvalue
  public var3=array(1,2,3);

its a wrong declaration Right is

public $var1="Test"; //String
public $var2=1.23; //Numericvalue
public $var3=array(1,2,3);

that's other wise every thing are fine . Thank you