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)
is_scalar — Comprueba si una variable es escalar
Descripción
Obtiene si la variable dada es escalar.
Las variables escalares son aquellas que contienen un integer, float, string o boolean. Tipos array, object y resource no son escalares.
Nota:
is_scalar() no tiene en cuenta resource como escalares ya que los recursos son tipos de datos abstractos que actualmente se basan en enteros. No debe confiar en este detalle de implementación, ya que puede cambiar.
Nota:
is_scalar() no considera que NULL sea escalar.
Parámetros
-
var -
La variable a ser evaluada.
Valores devueltos
Devuelve TRUE si var es escalar, FALSE
de lo contrario.
Ejemplos
Ejemplo #1 Ejemplo is_scalar()
<?php
function show_var($var)
{
if (is_scalar($var)) {
echo $var;
} else {
var_dump($var);
}
}
$pi = 3.1416;
$proteinas = array("hemoglobina", "citocromo c oxidasa", "ferredoxin");
show_var($pi);
show_var($proteinas)
?>
El resultado del ejemplo sería:
3.1416
array(3) {
[0]=>
string(10) "hemoglobina"
[1]=>
string(20) "citocromo c oxidasa"
[2]=>
string(10) "ferredoxin"
}
Ver también
- is_float() - Comprueba si el tipo de una variable es float
- is_int() - Comprueba si el tipo de una variable es integer
- is_numeric() - Comprueba si una variable es un número o un string numérico
- is_real() - Alias de is_float
- is_string() - Comprueba si una variable es de tipo string
- is_bool() - Comprueba si una variable es de tipo booleano
- is_object() - Comprueba si una variable es un objeto
- is_array() - Comprueba si una variable es un array
add a note
User Contributed Notes 4 notes
Dr K ¶
12 years ago
Anonymous ¶
11 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 ¶
12 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 ¶
13 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
