这比您想象的要复杂，因为它需要基数转换，并且基数转换是在整个数字上执行的，使用大整数运算。
这当然并不意味着我们不能专门为此目的创建所述大整数运算的有效实现。这是一个用零填充的实现（通常在 Java Card 上需要）并且不使用额外的内存（！）。如果你想保留它，你可能必须复制大端数的原始值 - 输入值被覆盖。强烈建议将其放入 RAM。
此代码只是将字节除以新的基数（小数为 10），然后返回余数。余数是下一个最低位。由于输入值现在已被除，下一个余数是仅比前一个位置高一个位置的数字。它一直在除法并返回余数，直到值为零并且计算完成。
算法的棘手部分是内部循环，它将值除以 10，同时使用字节尾除法返回余数。它每次运行提供一个余数/十进制数字。这也意味着函数的顺序是 O(n) 其中 n 是结果中的位数（将尾除定义为单个操作）。请注意，n 可以通过ceil(bigNumBytes * log_10(256)) 计算：其结果也存在于预先计算的BCD_SIZE_PER_BYTES 表中。 log_10(256) 当然是一个恒定的十进制值，高于2.408。
这是经过优化的最终代码（参见不同版本的编辑）：
/**
* 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 };
要扩展输入大小，只需计算下一个十进制大小并将其存储在表中...