php[tek] 2018 : Call for Speakers


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

is_scalarBir değişken sayıl mı diye bakar


bool is_scalar ( mixed $değişken )

Belirtilen değişkenin değeri sayıl türlerden biri ise TRUE döner.

Sayıl türler: integer, float, string ve boolean. Sayıl olmayan türler: array, object ve resource.


is_scalar() işlevi resource türündeki değerleri, tamsayılara dayalı soyut veri türleri olduklarından sayıl değerler olarak ele almaz. Bu gerçeklenim ayrıntısı bir kural olarak ele alınmamalıdır, ileride değişebilir.



Sınanacak değişken.

Dönen Değerler

Belirtilen değişkenin değeri sayıl türlerden biri ise TRUE, aksi takdirde FALSE döner.


Örnek 1 - is_scalar() örneği

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



Yukarıdaki örneğin çıktısı:

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

Ayrıca Bakınız

  • is_float() - Değişken float türünde mi diye bakar
  • is_int() - Değişken bir tamsayı mı diye bakar
  • is_numeric() - Değişken bir sayı veya bir sayısal dizge mi diye bakar
  • is_real() - is_float işlevinin takma adıdır
  • is_string() - Değişken string türünde mi diye bakar
  • is_bool() - Değişken boolean türünde mi diye bakar
  • is_object() - Değişken object türünde mi diye bakar
  • is_array() - Değişkenin bir dizi içerip içermediğine bakar

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User Contributed Notes 4 notes

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.

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
Dr K
12 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.
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.

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
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