将整数的二进制表示形式转换为 Java 卡中的 ASCII

Question

我想将以二进制格式表示的任意长度的整数转换为 ASCII 格式。

一个例子是整数33023，十六进制字节是0x80ff。我想将 0x80ff 表示为 33023 的 ASCII 格式，它具有 0x3333303233.

的十六进制表示

我正在使用无法识别 String 类型的 Java Card 环境，因此我必须通过二进制操作手动进行转换。

解决此问题的最有效方法是什么，因为 Java 16 位智能卡上的卡环境非常受限。

Answer 1

这比您想象的更棘手，因为它需要进行基数转换，并且使用大整数算法对整个数字执行基数转换。

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算法的有效实现。这是一个用零填充的实现（这通常在 Java 卡上是必需的）并且不使用额外的内存（！）。如果你想保留它，你可能必须复制 big endian 数字的原始值 - 输入值被覆盖。强烈建议将其放入 RAM。

此代码简单地将字节除以新基数（小数为 10），返回余数。余数是下一个最低数字。由于输入值现在已被除，下一个余数是比前一个数字高一个位的数字。它不断除法并返回余数，直到值为零，计算完成。

该算法的棘手部分是内部循环，它将值除以 10，同时使用字节尾部除法返回余数。它为每个运行提供一个余数/小数位。这也意味着函数的顺序是 O(n) 其中 n 是结果中的位数（将尾部除法定义为一次操作）。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 table 中。 log_10(256) 当然是一个常数十进制值，高于 2.408.

这是经过优化的最终代码（请参阅不同版本的编辑）：

/**
 * 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，只需计算下一个小数大小并将其存储在 table...

将整数的二进制表示形式转换为 Java 卡中的 ASCII

Converting binary representation of integers to ASCII in Java Card

java

binary

integer

javacard