这比你想象的更加棘手，因为它需要进行基数转换，并且基数转换是在整个数字上执行的，使用大整数算术。
当然，这并不意味着我们不能为此创建一种有效的大整数算术实现。这里提供了一个实现，它在左侧填充零（在Java Card上通常需要），并且不使用额外的内存(!)。但如果你想保存原始的大端数值，则可能需要复制它——输入值将被覆盖。将其放在RAM中是十分推荐的。
该代码只需将字节除以新的基数（10表示十进制），返回余数。余数就是下一个最低位数字。由于输入值现在已经被除以，所以下一个余数是比之前的数字位更重要的数字。它不断地除以并返回余数，直到值为零且计算完成。
算法的棘手部分是内部循环，它在原地将值除以10，同时使用尾部划分通过字节返回余数。它每次运行都提供一个余数 / 十进制数字。这也意味着该函数的顺序为O(n)，其中n是结果中的数字位数（将尾部划分定义为单个操作）。请注意，n可以通过ceil(bigNumBytes * log_10(256))计算得出：其结果也在预先计算的BCD_SIZE_PER_BYTES表中出现。当然，log_10(256)是一个常数十进制值，大约高达2.408。
以下是最终的代码优化（请参见编辑以获取不同版本）：
/**
* Converts an unsigned big endian value within the buffer to the same value
* stored using ASCII digits. The ASCII digits may be zero padded, depending
* on the value within the buffer.
* <p>
* <strong>Warning:</strong> this method zeros the value in the buffer that
* contains the original number. It is strongly recommended that the input
* value is in fast transient memory as it will be overwritten multiple
* times - until it is all zero.
* </p>
* <p>
* <strong>Warning:</strong> this method fails if not enough bytes are
* available in the output BCD buffer while destroying the input buffer.
* </p>
* <p>
* <strong>Warning:</strong> the big endian number can only occupy 16 bytes
* or less for this implementation.
* </p>
*
* @param uBigBuf
* the buffer containing the unsigned big endian number
* @param uBigOff
* the offset of the unsigned big endian number in the buffer
* @param uBigLen
* the length of the unsigned big endian number in the buffer
* @param decBuf
* the buffer that is to receive the BCD encoded number
* @param decOff
* the offset in the buffer to receive the BCD encoded number
* @return decLen, the length in the buffer of the received BCD encoded
* number
*/
public static short toDecimalASCII(byte[] uBigBuf, short uBigOff,
short uBigLen, byte[] decBuf, short decOff) {
// variables required to perform long division by 10 over bytes
// possible optimization: reuse remainder for dividend (yuk!)
short dividend, division, remainder;
// calculate stuff outside of loop
final short uBigEnd = (short) (uBigOff + uBigLen);
final short decDigits = BYTES_TO_DECIMAL_SIZE[uBigLen];
// --- basically perform division by 10 in a loop, storing the remainder
// traverse from right (least significant) to the left for the decimals
for (short decIndex = (short) (decOff + decDigits - 1); decIndex >= decOff; decIndex--) {
// --- the following code performs tail division by 10 over bytes
// clear remainder at the start of the division
remainder = 0;
// traverse from left (most significant) to the right for the input
for (short uBigIndex = uBigOff; uBigIndex < uBigEnd; uBigIndex++) {
// get rest of previous result times 256 (bytes are base 256)
// ... and add next positive byte value
// optimization: doing shift by 8 positions instead of mul.
dividend = (short) ((remainder << 8) + (uBigBuf[uBigIndex] & 0xFF));
// do the division
division = (short) (dividend / 10);
// optimization: perform the modular calculation using
// ... subtraction and multiplication
// ... instead of calculating the remainder directly
remainder = (short) (dividend - division * 10);
// store the result in place for the next iteration
uBigBuf[uBigIndex] = (byte) division;
}
// the remainder is what we were after
// add '0' value to create ASCII digits
decBuf[decIndex] = (byte) (remainder + '0');
}
return decDigits;
}
/*
* pre-calculated array storing the number of decimal digits for big endian
* encoded number with len bytes: ceil(len * log_10(256))
*/
private static final byte[] BYTES_TO_DECIMAL_SIZE = { 0, 3, 5, 8, 10, 13,
15, 17, 20, 22, 25, 27, 29, 32, 34, 37, 39 };
为了扩展输入尺寸，只需在表中计算并存储下一个十进制尺寸即可...
String，那么您是否至少有char数组和/或byte数组？ - Kevin Andersonshort[]是String的一个很好的替代品；即使是byte[]也可以用于ASCII。只要我们在short（或byte）上也有/（除法）和％（余数）运算符，我们就万事大吉（;->）。 - Kevin Andersonint和int[]通常不支持Java Card平台。在指定数字格式时，您只是展示了一个例子：0x80FF。这可能是一个由两个字节0x80和0xFF组成的无符号大端格式。您正在完全指定输出格式，请确保提供输入格式的说明。 - Maarten Bodewes