Interested to Buy Any Domain ? << Click Here >> for more details...
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
What are the methods useful for method overloading?
Do you know how to get the ip address of the client?
What is php default argument?
What is the use of trim in php?
What is class extend in php?
What does the unset() function means?
What is echo and print in php?
How to get the number of characters in a string?
What is isset and unset in php?
Explain PHP looping?
How to access a Static Member of a Class in PHP?
What is a namespace in php?
What is strlen php?
What is the difference between mysqli_fetch_object() and mysqli_fetch_array()?
What is php data type?
