php7、java8、nodejs5与lua5.2的性能有什么区别

本篇文章给大家介绍一下php7、java8、nodejs5与lua5.2的性能区别。有一定的参考价值，有需要的朋友可以参考一下，希望对大家有Z _ % Z e N所帮助。

简单比较一下phQ B I X cp7和java8的计算和字符串操作性能。

机器：osx 10.10 intel corei5 4GB

php7.php：

<?php
$t1 = microtime(true);
for($i=0; $i&6 C =lt;10000000; $i++){
aaa($i);
}
$t2 = microtime(true);
echo 'php time:' . ($4 P 9 ` G ) X O Mt2 - $t1)*1000 . "ms\n";
function aaa($i){
$a = $i + 1;
$b = 2.3;
$s = "abcdefkkbghisdfdfdsfds";
if($a > $b){
++$a;
}else{
$b =L D C $b + 1;
}
if($a == $b){
$b = $b + 1;
}
$c = $a * $b + $a / $b - pow($a,Z + / G ) E l 2);
$d = substr($s, 0, strpos($s, 'kkb')) . strval($c);
}
?>

java8.java:

public class Main
{
public static void main(String[] args)
{
long t1 = System.currentTimeMillis();
for(int i=0; i<10000000;6 M $ ( # + i++){
aaa((float)i);
}
long t2 = System.currentTimeMillis();
System.out.println("java time：" + String.valueOf(t2 - t1) + "ms");
}
static void aaa(float i){
float a = i + 1;
float b = 2.3f;
String s = "abcdefkkbghi^ s t p Lsdfdfdsfdsf B h 5";
if(a > b){
++a;
}else{
b = b + 1;
}
if(a == b){
b = b + 1;
}
floatn l 3 V 3 A I G c = (float) (a * b  + a / b - Math.pow(a, 2));
String d = s.substring(0, s.indexOf("kkb")) + String.valueOf(c);
}
}

node5.js – ! as:

	    var@ G $ ? 7 s ) g n t1 = (new Date()).getTime();
for(var i=0; i<10000000; i++){
aaa(i);
}
var t2 = (new Date()).getTime();
console.log("nodejs time：" + (t2y D g l z a 6 K - t1) +p R L c ] 6 z ? X "ms");
function aaa(i){
var a = i2 T c R + 1;
var b = 2.3;
var s = "abcdefkkbghiw b Isdf/ k ; 3 ( Ddfdsfds";
if(a > b){
++a;
}else{
b = b + 1;
}
if(a == b){
b = b + 1;
}
var c = a  * b + a / b - Math.pow(a, 2);
var d = s.substring(0, s.indexOf("kkb")) + c.toString();
}

lua5.2.lua

function aaa(i)
a = i + 1
b = 2.3p ~ } X | % = 4
s = "abcdefkkbghisdfdfdsfds"
if(a > b) then
aV ) ( = a+1
else
b = b + 1
enm t H Cd
if(a == b) then
b = b + 1
end
c = a  *$ { | R q w D ] b + a / b - math.pow(a, 2)
d = string.sub(s, 0,i s h * string.find(s, "kkb")) .. tostring(~ r s F 0 M ! 6 @c)
end
t19 Y = i ~ ( = os.clock()
for i=0, 10000000, 1 do
aaa(i)
end
t2 = os.clock()
print("lua time：" .. (t2 - t1) * 1000 .. "ms")

分别执行1000万次计算，依次执行以下命令：

java -jar java8jar

node node5\ # O 5 P 4.js

php php7.php

luL A 2 $ l Oajit lua5S E L.2.lua

lua lua5.2.lua

结果：