PHP 7.0.26 Released

is_scalar

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

is_scalar 変数がスカラかどうかを調べる

説明

bool is_scalar ( mixed $var )

指定した変数がスカラかどうかを調べます。

スカラ変数には integerfloatstring あるいは boolean が含まれます。 arrayobject および resource はスカラではありません。

注意:

リソース型は現在整数に基づく抽象型であるため、 is_scalar()resource 型の値を スカラ値と判定しません。この実装の詳細は変更される可能性があるため、 前堤にするべきではありません。

注意:

is_scalar() は、NULL をスカラとは見なしません。

パラメータ

var

評価する変数。

返り値

var がスカラの場合に TRUE、 それ以外の場合に FALSE を返します。

例1 is_scalar() の例

<?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)

?>

上の例の出力は以下となります。

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

参考

  • is_float() - 変数の型が float かどうか調べる
  • is_int() - 変数が整数型かどうかを検査する
  • is_numeric() - 変数が数字または数値形式の文字列であるかを調べる
  • is_real() - is_float のエイリアス
  • is_string() - 変数の型が文字列かどうかを調べる
  • is_bool() - 変数が boolean であるかを調べる
  • is_object() - 変数がオブジェクトかどうかを検査する
  • is_array() - 変数が配列かどうかを検査する

add a note add a note

User Contributed Notes 4 notes

up
10
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
up
8
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.
up
3
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.
up
-3
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
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