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

is_scalar Informa se é uma váriavel escalar


is_scalar ( mixed $var ) : bool

Verifica se a dada variável é uma escalar.

Variáveis escalares são as que contém integer, float, string ou boolean. os tipos array, object e resource não são escalares.


is_scalar() não considera o tipo resource como um valor escalar, apesar de tipos resouce ser uma abstração de dados, atualmente baseados em inteiros. Esse detalhe de implementação poderá ser modificado futuramente.


is_scalar() não considera NULL como sendo do tipo escalar.



A variável a ser avaliada.

Valor Retornado

Retorna true se var é um escalar, false caso contrário.


Exemplo #1 Exemplo da is_scalar()

function show_var($var)
    if (
is_scalar($var)) {
    } else {
$pi 3.1416;
$proteins = array("hemoglobin""cytochrome c oxidase""ferredoxin");



O exemplo acima irá imprimir:

array(3) {
  string(10) "hemoglobin"
  string(20) "cytochrome c oxidase"
  string(10) "ferredoxin"

Veja Também

  • is_float() - Informa se a variável é do tipo float
  • is_int() - Informa se a variável é do tipo inteiro
  • is_numeric() - Informa se a variável é um número ou uma string numérica
  • is_real() - Sinônimo de is_float
  • is_string() - Informa se a variável é do tipo string
  • is_bool() - Verifica se a variável é um boleano
  • is_object() - Informa se a variável é um objeto
  • is_array() - Verifica se a variável é um array

add a note add a note

User Contributed Notes 4 notes

Dr K
17 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.
16 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.

echo('is_scalar() test:'.EOL);
"NULL: "      . print_R(is_scalar(NULL),     true) . EOL);
"false: "    . print_R(is_scalar(false),   true) . EOL);
"(empty): "  . print_R(is_scalar(''),      true) . EOL);
"0: "         . print_R(is_scalar(0),       true) . EOL);
"'0': "      . print_R(is_scalar('0'),     true) . EOL);

is_scalar() test:
false: 1
(empty): 1
0: 1
'0': 1

   * NULL is NOT a scalar
   * false, (empty string), 0, and "0" ARE scalars
efelch at gmail dot com
17 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
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.

class objet_t {

$this->a = "patate_poil";

   function &
// 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);

"$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
To Top