just a warning as it appears that an empty value is not a scalar.
although the manual says scalars are:
-> integers (eg. 243)
-> floats (eg. 14.4)
-> strings (eg "this is a string")
-> booleans (eg true or false)
this might seem easy but rembere that empty values are:
-> NULL
-> false
-> empty string
-> 0
-> "0"
thus the true list of scalars must be:
-> integers large then zero
-> floats larger then zero
-> strings with a lenght large then zero
-> string that do not contain only the number 0
-> boolean value of true.
especialy the boolean value false can be tricky as true is
considerd to be scalar but false is not. or a string that is
initialized with a zero lenght here's an example.
$string = ""; //initialize our string.
function store_var($var)
{
if(is_scalar($var)){
print("our var is a scalar.");
}else{
print("you tried to store something else here.");
}
}
$this->store_var($string); // does not work as expected
this might be usefull when you create method to store a
variable inside a class but do not want to store other stuff
like object, arrays or resources, but storing an empy value
is oke. then is_scalar does not work as empty values are
incorrectly not seens as a bool, string, integer or float.
this is an important downside to is_scalar wich might cause
some people to loose there hair from headscrachting.
this might be added to the manual as a warning or something.
is_scalar
描述
bool is_scalar ( mixed var )如果给出的变量参数 var 是一个标量，is_scalar() 返回 TRUE，否则返回 FALSE。
标量变量是指那些包含了 integer、float、string 或 boolean的变量，而 array、object 和 resource 则不是标量。
<?php |
注: 尽管当前的 resource 类型是居于整数的，但 is_scalar() 不会把它们当作是标量，因为资源是抽象数据类型。不能依赖于执行细节，因为它可能会改变。
参见 is_bool()、is_numeric()、is_float()、is_int()、is_real()、is_string()、is_object()、is_array() 和 is_integer()。
add a note
User Contributed Notes
docey
21-May-2006 10:02
21-May-2006 10:02
efelch at gmail dot com
18-Aug-2005 05:31
18-Aug-2005 05:31
A scalar is a single item or value, compared to things like arrays and objects which have multiple values. This tends to be the standard definition of the word in terms of programming. An integer, character, etc are scalars. Strings are probably considered scalars since they only hold "one" value (the value represented by the characters represented) and nothing else.
Dr K
14-Aug-2005 06:48
14-Aug-2005 06:48
Having hunted around the manual, I've not found a clear statement of what makes a type "scalar" (e.g. if some future version of the language introduces a new kind of type, what criterion will decide if it's "scalar"? - that goes beyond just listing what's scalar in the current version.)
In other lanuages, it means "has ordering operators" - i.e. "less than" and friends.
It (-:currently:-) appears to have the same meaning in PHP.
popanowel HAT hotmailZ DOT cum
15-May-2004 03:31
15-May-2004 03:31
Hi ... for newbees here, I just want to mention that reference and scalar variable aren't the same. A reference is a pointer to a scalar, just like in C or C++.
<? php // simple reference to scalar
$a = 2;
$ref = & $a;
echo "$a <br> $ref";
?>
this should print out: "2 <br> 2".
Scalar class also exists. Look below:
<? php
class Object_t {
var $a;
function Object_t () // constructor
{
$this->a = 1;
}
}
$a = new Object_t; // we define a scalar object
$ref_a = &a;
echo "$a->a <br> $ref->a";
?>
again, this should echo: "1 <br> 1";
Here is another method isued in OOP to acheive on working only over reference to scalar object. Using this, you won't ever have to ask yourself if you work on a copy of the scalar or its reference. You will only possess reference to the scalar object. If you want to duplicate the scalar object, you will have to create a function for that purpose that would read by the reference the values and assign them to another scalar of the same type... or an other type, it is as you wish at that moment.
<?php
class objet_t {
var $a;
function object_t
{
$this->a = "patate_poil";
}
}
function &get_ref($object_type)
{
// here we create a scalar object in memory
// and we return it by reference to the calling
// control scope.
return &new $object_type;
}
$ref_object_t = get_ref(object_t);
echo "$ref_object_t->a <br>";
?>
this should echo: "patate_poit <br>".
The only thing that I try to demonstrate is that scalar variable ARE object in memory while a reference is usualy a variable (scalar object) that contain the address of another scalar object, which contain the informations you want by using the reference.
Good Luck!
otek is popanowel HAT hotmailZ DOT cum
bps7j at yahoNOSPAMo.com
09-Feb-2004 01:56
09-Feb-2004 01:56
is_scalar(null) is false. Apparently a variable needs to have a value to be considered a scalar.