时间复杂度或＆LT隐性成本;数组名称＆gt;。长度在java中 [英] time complexity or hidden cost of <Array Name>.length in java

问题描述

我看着在Java中的一个项目，发现上面写着象下面这样为循环：

 的for（int i = 1; I＆LT;则为a.length;我++）
{
    ...........
    ...........
    ...........
}
 

我的问题是：这是昂贵的计算则为a.length （这里是数组名）？如果没有那么如何则为a.length 是越来越内部计算（指JVM如何确保O（1）访问此）？是类似于：

  INT长度=则为a.length;
的for（int i = 1; I＆LT;长度;我+ +）
{
    ...........
    ...........
    ...........
}
 

即。像访问函数中的局部变量的值。谢谢你。

解决方案

  

我的问题是：这是昂贵的计算则为a.length

没有。这只是一个外地在阵列上（见 JLS节10.7 ）。这不是昂贵的，和JVM知道它永远不会改变，可以适当优化循环。 （事实上​​，我所期望的良好的JIT注意到初始化的变量与非负数的正常模式，检查它的小于长度，然后访问阵列 - 如果注意到，就可以删除数组边界检查。）

I was looking at a project in java and found a for loop which was written like below:

for(int i=1; i<a.length; i++)
{
    ...........
    ...........
    ...........
}

My question is: is it costly to calculate the a.length (here a is array name)? if no then how a.length is getting calculated internally (means how JVM make sure O(1) access to this)? Is is similar to:

int length = a.length;
for(int i=1; i<length; i++)
{
    ...........
    ...........
    ...........
}

i.e. like accessing a local variable's value inside the function. Thanks.

解决方案

My question is: is it costly to calculate the a.length

No. It's just a field on the array (see JLS section 10.7). It's not costly, and the JVM knows it will never change and can optimize loops appropriately. (Indeed, I would expect a good JIT to notice the normal pattern of initializing a variable with a non-negative number, check that it's less than length and then access the array - if it notices that, it can remove the array boundary check.)

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