is_scalar

(PHP 4 >= 4.0.5, PHP 5, PHP 7)

is_scalar — Determină dacă o variabilă este un scalar

Descrierea

is_scalar ( mixed $var ) : bool

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.

Notă:
is_scalar() nu consideră NULL ca scalar.

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"
}

A se vedea și

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

User Contributed Notes 4 notes

down

Dr K ¶

18 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.

down

Anonymous ¶

17 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

down

efelch at gmail dot com ¶

18 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.

down

-11

popanowel HAT hotmailZ DOT cum ¶

19 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

add a note