PHP 5.6.0RC3 is available

is_scalar

(PHP 4 >= 4.0.5, PHP 5)

is_scalar Finds whether a variable is a scalar

Description

bool is_scalar ( mixed $var )

Finds whether the given variable is a scalar.

Scalar variables are those containing an integer, float, string or boolean. Types array, object and resource are not scalar.

Note:

is_scalar() does not consider resource type values to be scalar as resources are abstract datatypes which are currently based on integers. This implementation detail should not be relied upon, as it may change.

Note:

is_scalar() does not consider NULL to be scalar.

Parameters

var

The variable being evaluated.

Return Values

Returns TRUE if var is a scalar FALSE otherwise.

Examples

Example #1 is_scalar() example

<?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)

?>

The above example will output:

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

See Also

  • is_float() - Finds whether the type of a variable is float
  • is_int() - Find whether the type of a variable is integer
  • is_numeric() - Finds whether a variable is a number or a numeric string
  • is_real() - Alias of is_float
  • is_string() - Find whether the type of a variable is string
  • is_bool() - Finds out whether a variable is a boolean
  • is_object() - Finds whether a variable is an object
  • is_array() - Finds whether a variable is an array

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User Contributed Notes 6 notes

up
8
Anonymous
8 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
6
Dr K
8 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
0
efelch at gmail dot com
8 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
-1
webmaster at oehoeboeroe dot nl
5 years ago
Here's a little function that will test whether a variable can be used as offset to an array.

<?php
function is_offset(&$var) {
    return (
is_scalar($var) || is_null($var)) && !is_resource($var);
}
?>

The resource check is currently redundant, but according to the manual that may change in the future.
up
-1
popanowel HAT hotmailZ DOT cum
10 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
up
-3
bps7j at yahoNOSPAMo.com
10 years ago
is_scalar(null) is false.  Apparently a variable needs to have a value to be considered a scalar.
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