is_scalar

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

is_scalar 변수가 스칼라인지 확인

설명

bool is_scalar ( mixed $var )

주어진 변수가 스칼라인지 확인합니다.

스칼라 변수는 integer, float, string, boolean을 포함합니다. array, object, resource는 스칼라가 아닙니다.

Note:

is_scalar()resource형 값을 스칼라로 간주하지 않습니다. 자원은 추상적인 자료형으로 현재는 정수에 기반하고 있지만, 이 구현 상세에 의존해선 안되며, 변경될 수 있습니다.

인수

var

평가할 변수.

반환값

var가 스칼라이면 TRUE, 아니면 FALSE를 반환합니다.

예제

Example #1 is_scalar() 예제

<?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)

?>

위 예제의 출력:

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

참고

  • is_float() - 변수의 자료형이 소수인지 확인합니다
  • is_int() - 변수의 자료형이 정수인지 확인합니다
  • is_numeric() - 변수가 수나 수 문자열인지 확인합니다
  • is_real() - 별칭: is_float
  • is_string() - 변수의 자료형이 문자열인지 확인합니다
  • is_bool() - 변수가 논리형인지 확인
  • is_object() - 변수가 객체인지 확인합니다
  • is_array() - 변수가 배열인지 확인

add a note add a note

User Contributed Notes 4 notes

up
13
Dr K
14 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
8
Anonymous
13 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
efelch at gmail dot com
14 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
-6
popanowel HAT hotmailZ DOT cum
15 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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