boost uint16的dynamic_bitset拷贝位

Question

我需要创建24位的set。 首先（0）位必须由bool设置。 和其他（1 - 23）我需要从uint32中复制第一位数值

是否可以使用dynamic_bitset来实现？

我的代码我试过，但错了：

typedef boost::dynamic_bitset<unsigned char> DataType; 
DataType bs(24, intValue); 
bs.set(0, booleanValue); 

Answer 1

只需左移：

DataType bs(24, intValue); 
    bs <<= 1; 
    bs.set(0, boolValue); 

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#include <boost/dynamic_bitset.hpp> 
#include <iostream> 
typedef boost::dynamic_bitset<unsigned char> DataType; 

int main() { 
    using namespace std; // for readability on SO 
    cout << hex << showbase; 

    uint32_t intValue = 0x666; 
    cout << "input: " << intValue; 

    DataType bs(24, intValue); 

    cout << "\n#1: " << bs << " " << bs.to_ulong(); 

    bs <<= 1; 
    cout << "\n#2: " << bs << " " << bs.to_ulong(); 

    bs.set(0, true); 

    cout << "\n#3: " << bs << " " << bs.to_ulong(); 
} 

打印：

input: 0x666 
#1: 000000000000011001100110 0x666 
#2: 000000000000110011001100 0xccc 
#3: 000000000000110011001101 0xccd 

Answer 2

所以，我设法做到这一点以这种方式，没有提升的bitset：

uint32_t buffer(0xAAAAAAAA); 
buffer = buffer << 1; 
buffer |= true << 0; 

unsigned char newRecord[3]; 

newRecord[0] = buffer; 
newRecord[1] = buffer << 8; 
newRecord[2] = buffer << 16;