PHP3 Manual | ||
---|---|---|
Prev | Chapter 6. Language constructs | Next |
A function may be defined using syntax such as the following:
function foo( $arg_1, $arg_2, ..., $arg_n ) { echo "Example function.\n"; return $retval; }
Any valid PHP3 code may appear inside a function, even other functions and class definitions.
Values are returned by using the optional return statement. Any type may be returned, including lists and objects.
function my_sqrt( $num ) { return $num * $num; } echo my_sqrt( 4 ); // outputs '16'.
Multiple values may not be returned, but the same effect can be achieved by returning a list:
function foo() { return array( 0, 1, 2 ); } list( $zero, $one, $two ) = foo();
Information may be passed to functions via the argument list, which is a comma-delimited list of variables and/or constants.
PHP3 supports passing arguments by value (the default), passing by reference, and default argument values. Variable-length argument lists are not supported, but a similar effect may be achieved by passing arrays.
By default, function arguments are passed by value. If you wish to allow a function to modify its arguments, you may pass them by reference.
If you wish a function's argument to always be passed by reference, you can prepend an ampersand (&) to the argument name in the function definition:
function foo( &$bar ) { $bar .= ' and something extra.'; } $str = 'This is a string, '; foo2( $str ); echo $str; // outputs 'This is a string, and something extra.'
If you wish to pass a variable by reference to a function which does not do this by default, you may prepend an ampersand to the argument name in the function call:
function foo( $bar ) { $bar .= ' and something extra.'; } $str = 'This is a string, '; foo2( $str ); echo $str; // outputs 'This is a string, ' foo2( &$str ); echo $str; // outputs 'This is a string, and something extra.'
A function may define C++-style default values for scalar arguments as follows:
function makecoffee( $type = "cappucino" ) { echo "Making a cup of $type.\n"; } echo makecoffee(); echo makecoffee( "espresso" );
The output from the above snippet is:
Making a cup of cappucino. Making a cup of espresso.
Note that when using default arguments, any defaults should be on the right side of any non-default arguments; otherwise, things will not work as expected. Consider the following code snippet:
function makeyogurt( $type = "acidophilus", $flavour ) { return "Making a bowl of $type $flavour.\n"; } echo makeyogurt( "raspberry" ); // won't work as expected
The output of the above example is:
Warning: Missing argument 2 in call to makeyogurt() in /usr/local/etc/httpd/htdocs/php3test/functest.html on line 41 Making a bowl of raspberry .
Now, compare the above with this:
function makeyogurt( $flavour, $type = "acidophilus" ) { return "Making a bowl of $type $flavour.\n"; } echo makeyogurt( "raspberry" ); // works as expected
The output of this example is:
Making a bowl of acidophilus raspberry.
Prev | Home | Next |
INCLUDE | Up | OLD_FUNCTION |