帮我解码出来 并说明方法 谢谢!!
.tip {
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`
end
}
.tip {
下面的东东看不看的懂?
begin 600
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`
end
}
/*
这是个简单的Base64编码程序
作者:Roc Chen [email protected]
Base64 使用US-ASCII子集的65个字符, 每个字符用6位表示
因此"m"的Base64值为38, 二进制形式是 100110. 对于文本串,编码过程如下。例如"men": 先转成US-ASCII值. "m"十进制 109
"e"十进制 101
"n"十进制 110 二进制 : m 01101101
e 01100101
n 01101110 三个8位连起来是24位
011011010110010101101110 然后分成4个6位
011011 010110 010101 101110 现在得到4个值,十进制为
27 22 21 46 对应的 Base64 字符是 :
b W V u 编码总是基于3个字符,从而产生4个Base64字符。
如果只剩1或2个字符,使用特殊字符"="补齐Base64的4字。
如,编码"me" 01101101 01100101
0110110101100101
011011 010110 0101
111111 (与,补足6位)
011011 010110 010100
b W U
b W U = ("=" 补足4字符) 于是 "bWU=" 就是"me"的Base64值. 再如编码 "m" 01101101
011011 01
111111
011011 010000
b Q = =
于是 "bQ==" 就是"m"的Base64值. 值得注意的是,MIME规定一行最多76个字符. */ static String BaseTable[] = {
"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P",
"Q","R","S","T","U","V","W","X","Y","Z","a","b","c","d","e","f",
"g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v",
"w","x","y","z","0","1","2","3","4","5","6","7","8","9","+","/"
};
public static void encode(String filename, BufferedWriter out) {
try {
File f = new File(filename);
FileInputStream fin = new FileInputStream(filename); // 读文件到BYTE数组
byte bytes[] = new byte[(int)(f.length())];
int n = fin.read(bytes); if (n < 1) return; // 没有内容 byte buf[] = new byte[4]; // base64 字符数组 int n3byt = n / 3; // 3 bytes 组数
int nrest = n % 3; // 分组后剩余 bytes
int k = n3byt * 3; //
int linelength = 0; // 行长
int i = 0; // 指针 // 3-bytes 分组 ...
while ( i < k ) {
buf[0] = (byte)(( bytes[i] & 0xFC) >> 2);
buf[1] = (byte)(((bytes[i] & 0x03) << 4) |
((bytes[i+1] & 0xF0) >> 4));
buf[2] = (byte)(((bytes[i+1] & 0x0F) << 2) |
((bytes[i+2] & 0xC0) >> 6));
buf[3] = (byte)( bytes[i+2] & 0x3F);
send(out, BaseTable[buf[0]]);
send(out, BaseTable[buf[1]]);
send(out, BaseTable[buf[2]]);
send(out, BaseTable[buf[3]]);
/*
以上代码可以优化,但会难以理解
buf[0]= (byte)(b[i] >> 2);
buf[1]= (byte)(((b[i] & 0x03) << 4)|(b[i+1]>> 4));
buf[2]= (byte)(((b[i+1] & 0x0F)<< 2)|(b[i+2]>> 6));
buf[3]= (byte)(b[i+2] & 0x3F);
send(out,BaseTable[buf[0]]+BaseTable[buf[1]]+
BaseTable[buf[2]]+BaseTable[buf[3]]);
*/ if ((linelength += 4) >= 76) {
send(out, "\r\n");
linelength = 0;
}
i += 3;
} // 处理尾部 ...
if (nrest==2) {
// 2 bytes left
buf[0] = (byte)(( bytes[k] & 0xFC) >> 2);
buf[1] = (byte)(((bytes[k] & 0x03) << 4) |
((bytes[k+1] & 0xF0) >> 4));
buf[2] = (byte)(( bytes[k+1] & 0x0F) << 2);
}
else if (nrest==1) {
// 1 byte left
buf[0] = (byte)((bytes[k] & 0xFC) >> 2);
buf[1] = (byte)((bytes[k] & 0x03) << 4);
}
if (nrest > 0) {
// 发送尾部
if ((linelength += 4) >= 76) send(out, "\r\n");
send(out, BaseTable[buf[0]]);
send(out, BaseTable[buf[1]]);
//
if (nrest==2) {
send(out, BaseTable[buf[2]]);
}
else {
send(out, "=");
}
send(out, "=");
}
out.flush();
//这里用到的send方法,请大家根据需要,自己写。可以是把结果输出到控制台,或发送邮件。
}
catch (Exception e) {
e.printStackTrace();
}
}
}