快速bitset追加？ [英] Fast bitset append?

问题描述

例如

$ b b

  char value = 31; 
 char n_bits = 5 
 fast_bitset bits; 
 bit.append（value，n_bits）; 
  

我已经尝试过boost :: dynamic_bitset和std :: vector。

$ b

旧帖 b $ b

我使用boost :: dynamic_bitset来打包一些数据。

通常我想一次打包〜5位，调用像：

  char value = 31; 
 char n_bits = 5 
 boost :: dynamic_bitset<>位; 
 for（char n = n_bits-1; n> = 0; --n）
 bits.push_back（（value>>> n）& 1）; 
  

但是这似乎对我来说效率不高，不能一次性添加所有的位？ / p>

例如

  char value = 31; 
 char n_bits = 5 
 boost :: dynamic_bitset<>位; 
 bit.append（value，n_bits）; 
  

我会期望它做类似的操作：

  template< typename T> 
 void append（T value，size_t n_bits）
 {
 value<< = sizeof（T） //使目标位最重要，删除所有非目标位。 
 
 // TODO：将位移入正确的位置，并使用OR将所有目标位一次性添加到bitset 
} 
  

为什么dynamic_bitset没有这样的函数？

解决方案

使用 resize 成员函数。

  / *将低位nbits位从值附加到设置。 * / 
 template< typename T> 
 void append（boost :: dynamic_bitset<& set，T value，size_t nbits）
 {
 set.resize（set.size（）+ nbits）; 
 for（size_t i = 1; i <= nbits; i ++）{
 set [set.size（） -  i] = value& 1; 
 value>> = 1; 
} 
} 
  

I'm looking for a bitset implementation with fast bit appending, where several bits can be efficiently appended in one go.

e.g.

char value = 31;
char n_bits = 5;
fast_bitset bits;
bits.append(value, n_bits);

I have so far tried boost::dynamic_bitset and std::vector. Both of which are slow.

---

Old Post

I'm using boost::dynamic_bitset to pack some data.

Usually I want to pack ~5 bits at a time which would result in a call like:

char value = 31;
char n_bits = 5;
boost::dynamic_bitset<> bits;
for(char n = n_bits-1; n >= 0; --n)
    bits.push_back((value >> n) & 1);

However this seems to me quite inefficient, couldn't one add all the bits in one go?

e.g.

char value = 31;
char n_bits = 5;
boost::dynamic_bitset<> bits;
bits.append(value, n_bits);

I would expect it to do something like:

template<typename T>
void append(T value, size_t n_bits)
{
     value <<= sizeof(T) - n_bits; // Make target bits most significant, remove all non-target bits.

     // TODO: Shift bits into proper place and use OR to add all target bits to bitset in one go
}

Why doesn't dynamic_bitset have such a function? Would it be possible to implement efficiently?

解决方案

Use the resize member function.

/* Append the lower-order nbits bits from value to set. */
template <typename T>
void append(boost::dynamic_bitset<> &set, T value, size_t nbits)
{
    set.resize(set.size() + nbits);
    for (size_t i=1; i<=nbits; i++) {
        set[set.size() - i] = value & 1;
        value >>= 1;
    }
}

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