c – 如何在没有未定义行为的情况下安全地偏移位？

这是我写的函数：

template <typename T,size_t count> static T convertBitSetToNumber( const std::bitset<count>& bitset )
{
    T result;
    #define targetSize (sizeof( T )*CHAR_BIT)
    if ( targetSize > count )
    {
        // if bitset is 0xF00,converting it as 0x0F00 will lose sign information (0xF00 is negative,while 0x0F00 is positive)
        // This is because sign bit is on the left.
        // then,we need to add a zero (4bits) on the right and then convert 0xF000,later,we will divide by 16 (2^4) to preserve sign and value

        size_t missingbits = targetSize - count;

        std::bitset<targetSize> extended;
        extended.reset(); // set all to 0
        for ( size_t i = 0; i != count; ++i )
        {
            if ( i < count )
                extended[i+missingbits] = bitset[i];
        }

        result = static_cast<T>( extended.to_ullong() );

        result = result >> missingbits;

        return result;
    }
    else
    {
        return static_cast<T>( bitset.to_ullong() );
    }
}

而“测试程序”：

uint16_t val1 = Base::BitsetUtl::convertBitSetToNumber<uint16_t,12>( std::bitset<12>( "100010011010" ) );
// val1 is 0x089A
int16_t val2 = Base::BitsetUtl::convertBitSetToNumber<int16_t,12>( std::bitset<12>( "100010011010" ) );
// val2 is 0xF89A

注意：请参阅与Ped7g的注释/交换,上面的代码是正确的并保留位符号,并对有符号或无符号位进行12-> 16位转换.但是如果你正在研究如何在签名对象上将0xABC0偏移到0x0ABC,那么答案可以帮到你,所以我不删除这个问题.

使用uint16作为目标类型时,请参阅该程序,如下所示：

uint16_t val = 0x89A0; // 1000100110100000
val = val >> 4;        // 0000100010011010

但是,使用int16_t时失败,因为0x89A0>> 4是0xF89A而不是预期的0x089A.

int16_t val = 0x89A0; // 1000100110100000
val = val >> 4;       // 2222200010011010

我不明白为什么>>运算符有时会插入0,有时为1.我无法找到如何安全地执行函数的最终操作(result = result>> missingbits;在某些时候一定是错的…)