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Php language.types.type juggling () function

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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.

= "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.

= "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:

= "abc"; // $a is a string
$a{1} = "f";   // $a is now "afc"

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

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.

= 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:

= (int) $bar;
$foo = ( int ) $bar;

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

= 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) {
"they are the same";

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

Php language.types.type juggling Function syntax tag

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Php language.types.type juggling syntax tutorial

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