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<Pseudo-types used in this documentationVariables>
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Last updated: Thu, 21 Aug 2003

Type Juggling

PHP does not require (or support) explicit type definition in variable declaration; a variable's type is determined by the context in which that variable is used. That is to say, if you assign a string value to variable $var, $var becomes a string. If you then assign an integer value to $var, it becomes an integer.

An example of PHP's automatic type conversion is the addition operator '+'. If any of the operands is a float, then all operands are evaluated as floats, and the result will be a float. Otherwise, the operands will be interpreted as integers, and the result will also be an integer. Note that this does NOT change the types of the operands themselves; the only change is in how the operands are evaluated.

<?php
$foo = "0";  // $foo is string (ASCII 48)
$foo += 2;   // $foo is now an integer (2)
$foo = $foo + 1.3;  // $foo is now a float (3.3)
$foo = 5 + "10 Little Piggies"; // $foo is integer (15)
$foo = 5 + "10 Small Pigs";     // $foo is integer (15)
?>

If the last two examples above seem odd, see String conversion to numbers.

If you wish to force a variable to be evaluated as a certain type, see the section on Type casting. If you wish to change the type of a variable, see settype().

If you would like to test any of the examples in this section, you can use the var_dump() function.

Note: The behaviour of an automatic conversion to array is currently undefined.

<?php
$a = "1";     // $a is a string
$a[0] = "f";  // What about string offsets? What happens?
?>

Since PHP (for historical reasons) supports indexing into strings via offsets using the same syntax as array indexing, the example above leads to a problem: should $a become an array with its first element being "f", or should "f" become the first character of the string $a?

The current versions of PHP interpret the second assignment as a string offset identification, so $a becomes "f", the result of this automatic conversion however should be considered undefined. PHP 4 introduced the new curly bracket syntax to access characters in string, use this syntax instead of the one presented above:

<?php
$a    = "abc"; // $a is a string
$a{1} = "f";   // $a is now "afc"
?>

See the section titled String access by character for more informaton.

Type Casting

Type casting in PHP works much as it does in C: the name of the desired type is written in parentheses before the variable which is to be cast.

<?php
$foo = 10;   // $foo is an integer
$bar = (boolean) $foo;   // $bar is a boolean
?>

The casts allowed are:

  • (int), (integer) - cast to integer

  • (bool), (boolean) - cast to boolean

  • (float), (double), (real) - cast to float

  • (string) - cast to string

  • (array) - cast to array

  • (object) - cast to object

Note that tabs and spaces are allowed inside the parentheses, so the following are functionally equivalent:

<?php
$foo = (int) $bar;
$foo = ( int ) $bar;
?>

Note: Instead of casting a variable to string, you can also enclose the variable in double quotes.

<?php
$foo = 10;            // $foo is an integer
$str = "$foo";        // $str is a string
$fst = (string) $foo; // $fst is also a string

// This prints out that "they are the same"
if ($fst === $str) {
    echo "they are the same";
}
?>

It may not be obvious exactly what will happen when casting between certain types. For more info, see these sections:



add a note add a note User Contributed Notes
Type Juggling
post_at_henribeige_dot_de
03-May-2003 11:37
If you want to do not only typecasting between basic data types but between classes, try this function. It converts any class into another. All variables that equal name in both classes will be copied.

function typecast($old_object, $new_classname) {
  if(class_exists($new_classname)) {
    $old_serialized_object = serialize($old_object);
    $new_serialized_object = 'O:' . strlen($new_classname) . ':"' . $new_classname . '":' .
                             substr($old_serialized_object, $old_serialized_object[2] + 7);
    return unserialize($new_serialized_object);
  }
  else
    return false;
}

Example:

class A {
  var $secret;
  function A($secret) {$this->secret = $secret;}
  function output() {echo("Secret class A: " . $this->secret);}
}

class B extends A {
  var $secret;
  function output() {echo("Secret class B: " . strrev($this->secret));}
}

$a = new A("Paranoia");
$b = typecast($a, "B");

$a->output();
$b->output();
echo("Classname \$a: " . get_class($a) . "Classname \$b: " . get_class($b));

Output of the example code above:

Secret class A: Paranoia
Secret class B: aionaraP
Classname $a: a
Classname $b: b
yury at krasu dot ru
27-Nov-2002 05:24
incremental operator ("++") doesn't make type conversion from boolean to int, and if an variable is boolean and equals TRUE than after ++ operation it remains as TRUE, so:

$a = TRUE;
echo ($a++).$a;  // prints "11"
29-Aug-2002 12:26
Printing or echoing a FALSE boolean value or a NULL value results in an empty string:
(string)TRUE //returns "1"
(string)FALSE //returns ""
echo TRUE; //prints "1"
echo FALSE; //prints nothing!
amittai at NOSPAMamittai dot com
21-Aug-2002 01:30
Uneven division of an integer variable by another integer variable will result in a float by automatic conversion -- you do not have to cast the variables to floats in order to avoid integer truncation (as you would in C, for example):

$dividend = 2;
$divisor = 3;
$quotient = $dividend/$divisor;
print $quotient; // 0.66666666666667

Amittai Aviram

<Pseudo-types used in this documentationVariables>
 Last updated: Thu, 21 Aug 2003
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