is_scalar

(PHP 4 >= 4.0.5, PHP 5, PHP 7, PHP 8)

is_scalar — 检测变量是否是一个标量

说明

is_scalar ( mixed $value ) : bool

检测变量是否为标量

标量变量是指 int、float、string 或 bool 类型的变量。 array、object 和 resource 类型不是标量。

Note:
is_scalar() 不会将 resource 类型值视为标量，因为当前 resources 是基于整数（integer）的抽象数据类型。不能依赖该执行细节，因为它可能会改变。

Note:
is_scalar() 不会将 NULL 检测为标量。

参数

value: 需要检测的变量。

返回值

如果 value 是标量，则返回 true ，否则返回 false 。

范例

Example #1 is_scalar() 示例


<?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)

?>

以上例程会输出：

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

参见

is_float() - 检测变量是否是浮点型
is_int() - 检测变量是否是整数
is_numeric() - 检测变量是否为数字或数字字符串
is_real() - is_float 的别名
is_string() - 检测变量是否是字符串
is_bool() - 检测变量是否是布尔值
is_object() - 检测变量是否是一个对象
is_array() - 检测变量是否是数组

User Contributed Notes

31-Jul-2006 08:59


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

efelch at gmail dot com 17-Aug-2005 02:31


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.

Dr K 14-Aug-2005 03:48


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.

popanowel HAT hotmailZ DOT cum 14-May-2004 12:31


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