如何找到无符号幅度和 2 的补码的最小位数？

Question

我对找到无符号大小和 2 的补码的最小位数有点困惑。到目前为止，这是我的推理：例如，

a) 243 decimal
Since 2^8 = 256, Unsigned and 2's compliment would both need a minimum 8 bits.
b) -56 decimal
This is impossible for unsigned.
2^6 = 64. One more bit is needed to show it is negative, so minimum 7 bits.

我的推理是否正确？

Answer 1

unsigned 的 "bits needed" 只是最高有效位（+1，取决于 MSB 的定义），对于二进制补码，您可以将该值取反并减一使其为正，然后为符号标志添加另一位。

int LeadingZeroCount(long value) {
    // http://en.wikipedia.org/wiki/Hamming_weight
    unsigned long x = value;
    x |= (x >> 1); x |= (x >> 2); x |= (x >> 4);
    x |= (x >> 8); x |= (x >> 16); x |= (x >> 32);
    x -= (x >> 1) & 0x5555555555555555;
    x = (x & 0x3333333333333333) + ((x >> 2) & 0x3333333333333333);
    x = (x + (x >> 4)) & 0x0F0F0F0F0F0F0F0F;
    x += x >> 8; x += x >> 16; x += x >> 32;
    return (sizeof(value) << 3) - (x & 0x7F);
}

int MostSignificantBit(long value) {
    return (sizeof(value) << 3) - LeadingZeroCount(value);
}

int BitsNeededUnsigned(unsigned long value) {
    return MostSignificantBit(value);
}

int BitsNeededTwosComplement(long value) {
    if (value < 0)
        return BitsNeededUnsigned(-value - 1) + 1;
    else
        return BitsNeededUnsigned(value);
}

int main() {
    printf("%d\n", BitsNeededUnsigned(243));
    printf("%d\n", BitsNeededTwosComplement(243));
    printf("%d\n", BitsNeededTwosComplement(-56));
    return 0;
}

至少，这是基于您对问题的定义。对我来说，+243 似乎需要 9 位作为二进制补码，因为符号位的 0 仍然相关。

如何找到无符号幅度和 2 的补码的最小位数？

How do I find the minimum number of bits for unsigned magnitude and 2's compliment?

binary

unsigned

bits

twos-complement