c++ - operator>> 和 << 输出在输出时为空

Question

我不明白为什么当我输出这两个函数时它们的输出都为空:

class uint128_t{

private:
    uint64_t UPPER, LOWER;

public:
    // constructors
    uint128_t(){
        UPPER = 0;
        LOWER = 0;
    }

    template <typename T>
    uint128_t(T rhs){
        UPPER = 0;
        LOWER = (uint64_t) rhs;
    }

    template <typename S, typename T>
    uint128_t(const S upper_rhs, const T lower_rhs){
        UPPER = (uint64_t) upper_rhs;
        LOWER = (uint64_t) lower_rhs;
    }

    uint128_t(const uint128_t & rhs){
        UPPER = rhs.UPPER;
        LOWER = rhs.LOWER;
    }

    //  RHS input args only

    // assignment operator
    template <typename T> uint128_t & operator=(T rhs){
        UPPER = 0;
        LOWER = (uint64_t) rhs;
        return *this;
    }

    uint128_t & operator=(uint128_t & rhs){
        UPPER = rhs.UPPER;
        LOWER = rhs.LOWER;
        return *this;
    }


    uint128_t operator<<(int shift){
        if (shift >= 128)
            return uint128_t(0, 0);
        else if (shift == 64)
            return uint128_t(LOWER, 0);
        else if (shift < 64)
            return uint128_t((UPPER << shift) + (LOWER >> (64 - shift)), LOWER << shift);
        else if ((128 > shift) && (shift > 64)){
            uint128_t a =uint128_t(LOWER << (shift - 64), 0);
            // a will show the correct values 
            std::cout << a.upper() << " " << a.lower() << std::endl;
            return uint128_t(LOWER << (shift - 64), 0);
            // in the program that includes this, printing out the values show 0 0
        }
    }

    uint128_t operator>>(int shift){
        if (shift >= 128)
            return uint128_t(0, 0);
        else if (shift == 64)
            return uint128_t(0, UPPER);
        else if (shift <= 64)
            return uint128_t(UPPER >> shift , ((UPPER << (64 - shift))) + (LOWER >> shift));
        else if ((128 > shift) && (shift > 64))
            return uint128_t(0, (UPPER >> (shift - 64)));
    }

    uint128_t operator<<=(int shift){
        *this = *this << shift;
        return *this;
    }

    uint128_t operator>>=(int shift){
        *this = *this >> shift;
        return *this;
    }

    const uint64_t upper() const {
        return UPPER;
    }

    const uint64_t lower() const {
        return LOWER;
    }

// lots of other stuff

};

int main(){

    uint128_t a(0x123456789abcdef1ULL, 0x123456789abcdef1ULL);

    a>>= 127; // or a <<= 127;
    std::cout <<a.upper() << " " <<a.lower() << std::endl;
    return 0;
}

http://ideone.com/jnZI9

谁能知道为什么？

Answer 1

您的 int 是 128 位，您正在将它向下(右)移动 127 位，这意味着最高位将移至最低位，所有其他位将为 0。

但是你的例子中的 int 是 0x1....，0x1(最高半字节)在二进制中是 0001，这确实没有设置高位。所以 0 是正确的输出。

如果您将 0x1... 更改为 0x8...(或任何高于 0x7 的值)，您很可能会在输出。