我有以下代码：
String curDir = "."; File fileObject = new File(curDir); File[] fileList = fileObject.listFiles(); float fileLengthMegabytes = (float)fileList[i].length() / 1000000;
方法fileList [i] .length()返回311字节作为Long类型.
3.88E-4
如何在fileLengthMegabytes变量中获得0,000311的预期输出？
解决方法
那就是科学符号.
而你得到388而不是311,因为你除以1000000而不是1048576(1024 * 1024)
编辑：311即使使用1048576也没有实现,这样你就可以得到370 …所以错误可能在你的calc中);
作为described here,您只需要通过格式化器将您的科学符号转换为十进制符号.
DecimalFormat df = new DecimalFormat("#.########");
return df.format(fileLengthMegabytes);
import java.util.*;
import java.lang.*;
import java.text.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
DecimalFormat df = new DecimalFormat("#.##########");
float fileLengthMegabytes1 = (float) 388 / 1000000;
float fileLengthMegabytes2 = (float) 388 / 1048576;
System.out.println("MB1 in Scientific Notation: " +
fileLengthMegabytes1);
System.out.println("MB1 in Decimal Notation: " +
df.format(fileLengthMegabytes1));
System.out.println("MB2 in Scientific Notation: " +
fileLengthMegabytes2);
System.out.println("MB2 in Decimal Notation: " +
df.format(fileLengthMegabytes2));
}
}
输出：
MB1 in Scientific Notation: 3.88E-4
MB1 in Decimal Notation: 0.000388
MB2 in Scientific Notation: 3.7002563E-4
MB2 in Decimal Notation: 0.0003700256
