Question

给定一个整数值，我需要一些方法来找出存储该值所需的最小字节数。该值可以是有符号或无符号，最多为64位。还要将符号位考虑为有符号整数。

例如：

                          8     requires 1 byte at minimum
               unsigned 255     requires 1 byte at minimum
                 signed 255     requires 2 bytes at minimum
                       4351     requires 2 bytes at minimum
                -4294967296     requires 5 bytes at minimum
unsigned 0xFFFFFFFFFFFFFFFF     requires 8 bytes at minimum

我可以想到一种快速而肮脏的方法来解决这个问题，使用许多if语句，但可能有更好的方法（例如更简单，更聪明，更快）。您可以假定使用签名int (long value, bool signed)的方法或两种方法int (long value)（用于签名）和int (ulong value)（用于无符号）。

Answer 1

让我试一试自己的问题。据我所知，这是一个正确的解决方案，但它在速度，简洁方面可能不是最佳的：

public static int GetMinByteSize(long value, bool signed)
{
    ulong v = (ulong)value;
    // Invert the value when it is negative.
    if (signed && value < 0)
        v = ~v;
    // The minimum length is 1.
    int length = 1;
    // Is there any bit set in the upper half?
    // Move them to the lower half and try again.
    if ((v & 0xFFFFFFFF00000000) != 0)
    {
        length += 4;
        v >>= 32;
    }
    if ((v & 0xFFFF0000) != 0)
    {
        length += 2;
        v >>= 16;
    }
    if ((v & 0xFF00) != 0)
    {
        length += 1;
        v >>= 8;
    }
    // We have at most 8 bits left.
    // Is the most significant bit set (or cleared for a negative number),
    // then we need an extra byte for the sign bit.
    if (signed && (v & 0x80) != 0)
        length++;
    return length;
}

Answer 2

static int BytesForNum(long value, bool signed)
{
    if (value == 0)
        return 1;
    if (signed)
    {
        if (value < 0)
            return CalcBytes(2 * (-1-value));
        else
            return CalcBytes(2 * value);
    }
    else
    {
        if (value < 0)
            throw new ArgumentException("Can't represent a negative unsigned number", "value");
        return CalcBytes(value);
    }
}
//should only be called with positive numbers
private static int CalcBytes(long value)
{
    int bitLength = 0;
    while (value > 0)
    {
        bitLength++;
        value >>= 1;
    }
    return (int)(Math.Ceiling(bitLength * 1.0 / 8));
}

我可能没有完全正确的签名代码，但这是一般的想法。

可包含整数值的最小字节数

2 个答案: