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is_scalar

(PHP 4 >= 4.0.5, PHP 5)

is_scalar Determină dacă o variabilă este un scalar

Descrierea

bool is_scalar ( mixed $var )

Determină dacă variabila dată este un scalar.

Variabilele scalare sunt acelea ce conțin un integer, float, string sau boolean. Tipurile array, object și resource nu sunt scalare.

Notă:

is_scalar() nu consideră valorile de tip resource a fi scalare deoarece resursele sunt tipuri de date abstracte care în prezent se bazează pe numere întregi. Nu trebuie de bazat pe acest detaliu de implementare, deoarece poate fi schimbat în viitor.

Parametri

var

Variabila ce este evaluată.

Valorile întoarse

Întoarce TRUE dacă var este un scalar, FALSE în caz contrar.

Exemple

Example #1 Exemplu 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)

?>

Exemplul de mai sus va afișa:

3.1416
array(3) {
  [0]=>
  string(10) "hemoglobin"
  [1]=>
  string(20) "cytochrome c oxidase"
  [2]=>
  string(10) "ferredoxin"
}

Vedeți de asemenea

  • is_float() - Determină dacă tipul unei variabile este float
  • is_int() - Determină dacă tipul unei variabile este integer
  • is_numeric() - Determină dacă o variabilă este un număr sau un string numeric
  • is_real() - Pseudonim pentru is_float
  • is_string() - Determină dacă tipul variabilei este string
  • is_bool() - Determină dacă o variablă este un boolean
  • is_object() - Determină dacă o variabilă este un obiect
  • is_array() - Determină dacă o variabilă este un array

add a note add a note

User Contributed Notes 6 notes

up
8
Anonymous
8 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
up
5
Dr K
9 years ago
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.
up
1
efelch at gmail dot com
9 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.
up
-2
webmaster at oehoeboeroe dot nl
5 years ago
Here's a little function that will test whether a variable can be used as offset to an array.

<?php
function is_offset(&$var) {
    return (
is_scalar($var) || is_null($var)) && !is_resource($var);
}
?>

The resource check is currently redundant, but according to the manual that may change in the future.
up
-2
popanowel HAT hotmailZ DOT cum
10 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
up
-6
bps7j at yahoNOSPAMo.com
10 years ago
is_scalar(null) is false.  Apparently a variable needs to have a value to be considered a scalar.
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