在PHP中找出一个数字/变量是奇数还是偶数的最简单最基本的方法是什么? 这和mod有关吗?

我试过一些剧本,但是…谷歌目前没有发送。


当前回答

试试这个,

$number = 10;
 switch ($number%2)
 {
 case 0:
 echo "It's even";
 break;
 default:
 echo "It's odd";
 }

其他回答

我做了一些测试,发现在mod, is_int和&-操作符之间,mod是最快的,紧随其后的是&-操作符。 Is_int几乎比mod慢4倍。

I used the following code for testing purposes: $number = 13; $before = microtime(true); for ($i=0; $i<100000; $i++) { $test = ($number%2?true:false); } $after = microtime(true); echo $after-$before." seconds mod<br>"; $before = microtime(true); for ($i=0; $i<100000; $i++) { $test = (!is_int($number/2)?true:false); } $after = microtime(true); echo $after-$before." seconds is_int<br>"; $before = microtime(true); for ($i=0; $i<100000; $i++) { $test = ($number&1?true:false); } $after = microtime(true); echo $after-$before." seconds & operator<br>"; The results I got were pretty consistent. Here's a sample: 0.041879177093506 seconds mod 0.15969395637512 seconds is_int 0.044223070144653 seconds & operator

//checking even and odd
$num =14;

$even = ($num % 2 == 0);
$odd = ($num % 2 != 0);

if($even){
    echo "Number is even.";
} else {
    echo "Number is odd.";
}
<?php
// Recursive function to check whether
// the number is Even or Odd 
function check($number){
    if($number == 0)
        return 1;
    else if($number == 1)
        return 0;
    else if($number<0)
        return check(-$number);
    else
        return check($number-2);        
}
  
// Check the number odd or even
$number = 35;
if(check($number))
    echo "Even";
else
    echo "Odd";
?>

因此,输出将是Odd

另一种方法是检查最后一位是否为偶数:

$value = "1024";// A Number
$even = array(0, 2, 4, 6, 8);
if(in_array(substr($value, -1),$even)){
  // Even Number
}else{
  // Odd Number
}

或者为了更快,使用isset()代替array_search:

$value = "1024";// A Number
$even = array(0 => 1, 2 => 1, 4 => 1, 6 => 1, 8 => 1);
if(isset($even[substr($value, -1)]){
  // Even Number
}else{
  // Odd Number
}

或者让它更快(有时胜过mod操作符):

$even = array(0, 2, 4, 6, 8);
if(in_array(substr($number, -1),$even)){
  // Even Number
}else{
  // Odd Number
}

这是一个时间测试来证明我的发现。

使用位操作: 在此方法中,您将找到带有1的数字的逐位与。如果位与为1,则该数字为奇数,否则为偶数。

function odd($var){
    return $var & 1;
}
function even($var){
    return !($var&1);
}
$arr = range(1,10);

echo "Odd Numbers: ";
echo "<pre>";
    print_r(array_filter($arr2, 'odd'));
echo "<pre>";

echo "<br>Even Numbers: ";
echo "<pre>";
    print_r(array_filter($arr2, 'even'));
echo "</pre>";

输出将是:

Odd Numbers:
Array
(
    [0] => 1
    [2] => 3
    [4] => 5
    [6] => 7
    [8] => 9
)

Even Numbers: 
Array
(
    [1] => 2
    [3] => 4
    [5] => 6
    [7] => 8
    [9] => 10
)