Java的内存复制(主要是数组)效率比较
今天在群里看到有一个人问怎么截取char[]子串,可以调用api也可以手动复制.以前也没有太关注效率问题,今天特意测试了一下,下面是结果:
public class ArrayCopy
{
public static void main(String[] args)
{
int size =1024*1024*5;
int i,length = size;
char oldArray[] = new char[size];
char newArray[] = new char[size];
for(i=0;i<size;i++)
oldArray[i]=(char) i;
long startTime = System.currentTimeMillis();
for(i=0;i<1000;i++)
{
selfCopy(oldArray, 0, newArray, 0, length);//100次2475 1000次19290
//System.arraycopy(oldArray, 0, newArray, 0, length);//940 10225
//newArray=oldArray.clone();// 2545 25425
//selfCopy(newArray, 0, newArray, 1, length-1);
//System.arraycopy(newArray, 0, newArray, 1, length-1);
}
long endTime = System.currentTimeMillis();
System.out.println("消耗时间: " + (endTime - startTime)+" ms.");
System.out.println(new String(oldArray).substring(100,200)+"\n\n"+new String(newArray).substring(100,200));
}
private static void selfCopyObj(Object oldArray, int startPosOld, Object newArray, int startPosNew, int length)
{
for (int i = 0; i < length; i++)
{
( (Object[] )newArray)[startPosNew++] = ( (Object[]) oldArray)[startPosOld++];
}
}
private static void selfCopy(char[] oldArray, int startPosOld, char[] newArray, int startPosNew, int length)
{
for (int i = 0; i < length; i++)
{
newArray[startPosNew++] = oldArray[startPosOld++];
}
}
}
[解决办法]
额，我说错了，我觉得内存块对拷不用“都得一个一个搬”跟多少位也没啥关系，只是内存之间的数据传递，又不通过数据总线。
