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 double, then all operands are evaluated as doubles, and the result will be a double. 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.

$foo = "0";  // $foo is a string (ASCII 48)
$foo++;      // $foo is the string "1" (ASCII 49)
$foo += 1;   // $foo is now an integer (2)
$foo = $foo + 1.3;  // $foo is now a double (3.3)
$foo = 5 + "10 Little Piggies";    // $foo is a double (15)
$foo = 5 + "10 Small Pigs";   // $foo is an integer (15)
     

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

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().

Determining variable types

Because PHP determines the types of variables and converts them (generally) as needed, it is not always obvious what type a given variable is at any one time. PHP includes several functions which find out what type a variable is. They are gettype(), is_long(), is_double(), is_string(), is_array(), and is_object().

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.

$foo = 10;   // $foo is an integer
$bar = (double) $foo;   // $bar is a double
      

The casts allowed are:

• (int), (integer) - cast to integer

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

• (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:

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

String conversion

When a string is evaluated as a numeric value, the resulting value and type are determined as follows.

The string will evaluate as a double if it contains any of the characters '.', 'e', or 'E'. Otherwise, it will evaluate as an integer.

The value is given by the initial portion of the string. If the string starts with valid numeric data, this will be the value used. Otherwise, the value will be 0 (zero). Valid numeric data is an optional sign, followed by one or more digits (optionally containing a decimal point), followed by an optional exponent. The exponent is an 'e' or 'E' followed by one or more digits.

$foo = 1 + "10.5";      // $foo is a double (11.5)
$foo = 1 + "-1.3e3";    // $foo is a double (-1299)
$foo = 1 + "bob-1.3e3"; // $foo is a double (1)
$foo = 1 + "bob3";      // $foo is an integer (1)
$foo = 1 + "10 Small Pigs";     // $foo is an integer (11)
$foo = 1 + "10 Little Piggies"; // $foo is a double (11); the string contains 'e'
      

For more information on this conversion, see the Unix manual page for strtod(3).