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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)
<!-- bad example, no real operator (must be used with variable, modifies it too)
$foo++;      // $foo is the string "1" (ASCII 49)
-->
$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)
<!--

TODO: explain ++/- - behaviour with strings

examples:

++'001' = '002'
++'abc' = 'abd'
++'xyz' = 'xza'
++'9.9' = '9.0'
++'-3'  = '-4'
- -'9'   = 8 (integer!)
- -'5.5' = '5.5'
- -'-9'  = -10 (integer)
- -'09'  = 8 (integer)
- -'abc' = 'abc'

-->
?>

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:

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