Explain about Type Juggling in PHP?

Answer Posted / rakesh kumar nautiyal

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"
?>

Is This Answer Correct ?    4 Yes 3 No



Post New Answer       View All Answers


Please Help Members By Posting Answers For Below Questions

What is php? Why it is used?

778


What is var_dump function in php?

791


What is the meaning of xdebug?

757


Do you know what is the use of rand() in php?

759


Why do we need abstract class in php?

714


Is php a backend?

735


How do functions work?

757


"mysql_fetch_row — Get a result row as an enumerated array",this sentence comes from the PHP offical manual.However ,i can not understand the words "enumerated array".I need some help.Thanks a lot to everyone that reply.

1745


How to fix "headers already sent" error in php

750


What is Gd PHP?

771


What is lamp in php?

732


Is php free to use?

750


Does php need to be installed?

726


What is super keyword in c++?

773


What types of Data Can Be Used as Array Keys?

774