java新手问题（关于四舍五入，保留小数问题），先谢谢各位了？？？？？？？

[lenoli]

double dC = 1000.2356
要把dC保留两位小数，同时四舍五入,怎么写代码？

[cold_blooded]

public class Test {

public static void main(String[] args) {
float[] float_args = {12345.26f, 1234.22f, 123.22f, 0.123456789f};
for (int i = 0; i < float_args.length; i++) {
System.out.println(floatFormat(float_args[i]));
}

double[] double_args = {12345.26d, 1234.22d, 123.22d, 0.123456789d};
for (int i = 0; i < double_args.length; i++) {
System.out.println(doubleFormat(double_args[i]));
}

}

public static String floatFormat(float f) {
DecimalFormat df = new DecimalFormat("0.#");
return df.format(f);
}

public static String doubleFormat(double d) {
DecimalFormat df = new DecimalFormat("0.##");
return df.format(d);
}
}

[manbaum]

double dC = 1000.2356
double dC1 = (double)((int)(dC * 100) + 0.5))/100.0;

[lenoli]

return Double.valueOf(new java.text.DecimalFormat(temp).format(v));

编译不过去阿

[micker]

double dC = 1000.2356;
dC=(double)Math.round(dC*100)/100;
System.out.print(dC);

[flyazdream]

/**

* 提供小数位四舍五入处理。

* @param v 需要四舍五入的数字

* @param scale 小数点后保留几位

* @return 四舍五入后的结果

*/

public static double round(double v,int scale){
String temp="#,##0.";
for (int i=0;i<scale ;i++ )
{
temp+="0";
}
return Double.valueOf(new java.text.DecimalFormat(temp).format(v));
}


Java带的Math.round不提供精确到位数。上面方法可以完成此功能

[flyazdream]

http://blog.csdn.net/faintbear/archive/2004/08/30/88657.aspx

[YuLimin]

1位小数就是Math.round(XX*10)/10

2位小数就是Math.round(XX*100)/100