求个进制转换方法解决方案

求个进制转换方法
将十进制的88转为16进制，然后放到byte数组中，入： byte[] bytes = new byte[]{0x00,0x00,0x00,0x58};

帮忙写个这样的算法，要求byte[]的大小为4位。

------解决方案--------------------

Java code


    public static byte[] getByte(int a)
    {
        byte[] bs = new byte[4];
        bs[0] = (byte)((a >> 24) & 0xff);
        bs[1] = (byte)((a >> 16) & 0xff);
        bs[2] = (byte)((a >> 8) & 0xff);
        bs[3] = (byte)((a) & 0xff);
        return bs;
    }

------解决方案--------------------
int i = 88;
String i16 = Integer.toHexString(i);
byte[] buf = i16.getBytes();
for(int j = 0;j < buf.length;j++){
   System.out.println(buf[j]);
}

------解决方案--------------------

Java code

    public static byte[] getByte(int a)
        {
            byte[] bs = new byte[4];
            bs[0] = (byte)((a >> 24) & 0xff);
            bs[1] = (byte)((a >> 16) & 0xff);
            bs[2] = (byte)((a >> 8) & 0xff);
            bs[3] = (byte)((a) & 0xff);
        return bs;
    }

------解决方案--------------------

探讨

不行的啊你试试

引用:

Java code

public static byte[] getByte(int a)
{
byte[] bs = new byte[4];
bs[0] = (byte)((a >> 24) &amp; 0xff);
bs[1] = (byte)((a >> 16) &amp; 0xff);
bs[2] ……

求个进制转换方法解决方案

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