java中<Array Name>.length的时间复杂度或隐藏成本 [英] time complexity or hidden cost of <Array Name>.length in java

问题描述

我在看一个 java 项目，发现了一个 for 循环，它的写法如下:

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++)
{
    ...........
    ...........
    ...........
}

我的问题是:计算 a.length(这里 a 是数组名)的成​​本高吗?如果没有，那么 a.length 是如何在内部计算的(意味着 JVM 如何确保 O(1) 访问这个)?类似于:

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++)
{
    ...........
    ...........
    ...........
}

即就像在函数内部访问局部变量的值一样.谢谢.

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

推荐答案

我的问题是:计算 a.length 的成本高吗

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

没有.它只是数组上的一个字段(参见 JLS 第 10.7 节).它并不昂贵，并且 JVM 知道它永远不会改变并且可以适当地优化循环.(确实，我希望一个好的 JIT 注意到用非负数初始化变量的正常模式，检查它是否小于 length 然后访问数组 - 如果它注意到，它可以去掉数组边界检查.)

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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