Question

有什么办法可以将两个8位位组分配给一个16位位组 e.g。

 bitset<8> first = var1;
 bitset<8> second = var2;
 bitset<16> = first+second;

Answer 1

如果您希望序列化它们：

bitset<16> result(first.to_ulong() * 0x100 + second.to_ulong());

首先是最重要的，第二个是最不重要的位。

Answer 2

将第一位大小移位。

这假设first.size（）+ second.size（）在c ++ 11中不比ullong大。

bitset<first.size()+second.size()> third(first.to_ullong() << first.size() | second.to_ullong());

Answer 3

如果您不限制使用bitset＆lt;＆gt;，您可能会发现逐位运算符最适合此类操作。当然，它更简洁，可能适合也可能不适合您自己的喜好。如，

unsigned bit_1 = 0xff,   // low order 8 bits set
         bit_2 = 0xff00, // high order 8 bits set
         bit_3 = 0,
         bit_4 = 0x1ce9;

bit_3 = bit_1 | bit_2;  // now bit_3 has all 16 bits set
bit_3 ^= bit_3;         // every bit that was set is now cleared
bit_3 = bit_4 & bit_1;  // using bit_1 as a mask, bit_3 now contains just the
                        // lower order bits that are set in bit_4, 0xe9

将两个8位位组分配给一个16位位组

3 个答案: