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.
is_scalar
(PHP 4 >= 4.0.5, PHP 5, PHP 7, PHP 8)
is_scalar — Verifica se la variabile è di tipo scalare
Descrizione
Verifica se la data variabile sia di tipo scalare
Le variabili scalari sono quelle contenenti valori di tipo integer, float, string oppure boolean. I tipi array, object e resource non sono scalari.
Nota:
La funzione is_scalar() non considera il tipo resource come valore scalare, dato che il tipo resource è una tipologia di dato astratto che attualmente si basa su interi. Tuttavia non ci si può basare su questo tipo di implementazione, in futuro potrebbe cambiare.
Nota:
is_scalar() non considera NULL come valore scalare.
Elenco dei parametri
-
var -
La variabile da valutare.
Valori restituiti
Restituisce true se la variabile indicata dal parametro var è di tipo scalare,
in caso contrario restituisce false.
Esempi
Example #1 Esempio di uso di is_scalar()
<?php
function show_var($var)
{
if (is_scalar($var)) {
echo $var;
} else {
var_dump($var);
}
}
$pi = 3.1416;
$proteins = array("hemoglobin", "cytochrome c oxidase", "ferredoxin");
show_var($pi);
show_var($proteins)
?>Il precedente esempio visualizzerà:
3.1416
array(3) {
[0]=>
string(10) "hemoglobin"
[1]=>
string(20) "cytochrome c oxidase"
[2]=>
string(10) "ferredoxin"
}
Vedere anche:
- is_float() - Verifica se una variabile è di tipo float (decimale a virgola mobile)
- is_int() - Verifica se una variabile è di tipo integer
- is_numeric() - Verifica se una variabile è un numero o una stringa numerica
- is_real() - Alias di is_float
- is_string() - Verifica se il tipo di una variabile sia stringa
- is_bool() - Verifica se una variabile è di tipo boolean
- is_object() - Verifica se una variabile è un object
- is_array() - Verifica se una variabile è un array
add a note
User Contributed Notes 4 notes
Dr K ¶
19 years ago
Anonymous ¶
18 years ago
Another warning in response to the previous note:
> just a warning as it appears that an empty value is not a scalar.
That statement is wrong--or, at least, has been fixed with a later revision than the one tested. The following code generated the following output on PHP 4.3.9.
CODE:
<?php
echo('is_scalar() test:'.EOL);
echo("NULL: " . print_R(is_scalar(NULL), true) . EOL);
echo("false: " . print_R(is_scalar(false), true) . EOL);
echo("(empty): " . print_R(is_scalar(''), true) . EOL);
echo("0: " . print_R(is_scalar(0), true) . EOL);
echo("'0': " . print_R(is_scalar('0'), true) . EOL);
?>
OUTPUT:
is_scalar() test:
NULL:
false: 1
(empty): 1
0: 1
'0': 1
THUS:
* NULL is NOT a scalar
* false, (empty string), 0, and "0" ARE scalars
efelch at gmail dot com ¶
19 years ago
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.
popanowel HAT hotmailZ DOT cum ¶
20 years ago
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
