二进制字符数组到 stringstream 并从缓冲区弹出

我有 20 字节的二进制字符数组。我想分成3个部分：4byte、8byte、8byte。我像下面这样实现它。它有效，但似乎我可以使用缓冲区流。我想知道如何使用它。

现在

void main()
{
    // _data is 20byte binary char array. 0000000000000000000000000000000000000000000001111001110001111111001110000010110000001011101101000000000000000000000000000000000000000000000000000000000000000001

    // strA (4 byte)
    string strA;
    for (std::size_t i = 0; i < 4; ++i) {
        strA += bitset<8>(_data.c_str()[i]).to_string();
    }

    cout << strA << endl; // 00000000000000000000000000000000

    // strB (8 byte)
    string strB;
    for (std::size_t i = 4; i < 12; ++i) {
        strB += bitset<8>(_data.c_str()[i]).to_string();
    }

    cout << strB << endl; // 0000000000000111100111000111111100111000001011000000101110110100

    // strC (8 byte)
    string strC;
    for (std::size_t i = 12; i < 20; ++i) {
        strC += bitset<8>(_data.c_str()[i]).to_string();
    }

    cout << strC << endl; // 0000000000000000000000000000000000000000000000000000000000000001
}

期待

我想这样实现。

void main()
{
    stringstream ss = _data;
    strA = ss.pop(4);
    strB = ss.pop(8);
    strC = ss.pop(8);
}

更新 1

感谢你们。我正在一一尝试你给我的所有答案。我是 C++ 的新手，所以需要时间来理解它。以下是 Anders K 的一篇。

struct S { char four[4]; char eight1[8]; char eight2[8]; }; 
struct S *p = reinterpret_cast<S*>(&_data); 
cout << p->four << endl; // => Output "(" I think I can find way to output

更新 2

它使用 string::substr 工作。谢谢扎基尔。

int main()
{
    // I don't know how to change to string value in smart way.. 
    string str;
    for (std::size_t i = 0; i < _data.size(); ++i) {
        str += bitset<8>(_data.c_str()[i]).to_string();
    }

    cout << str << endl; // 0000000000000000000000000000000000000000000001111001110001111111001110000010110000001011101101000000000000000000000000000000000000000000000000000000000000000001

    std::string d = str; // Your binary stream goes here
    int lenA = (4*8);  // First  4 Bytes
    int lenB = (8*8);  // Second 8 Bytes
    int lenC = (8*8);  // Last   8 Bytes

    std::string strA = d.substr(0,    lenA);
    std::string strB = d.substr(lenA + 1, lenB - 1);
    std::string strC = d.substr(lenA + lenB + 1, lenC - 1);

    cout << strA << endl; // 00000000000000000000000000000000
    cout << strB << endl; // 000000000000111100111000111111100111000001011000000101110110100
    cout << strC << endl; // 000000000000000000000000000000000000000000000000000000000000001
}

更新 3

当我尝试 Scheff 的方式时出现错误。这是我的错，我想我可以解决它。我认为我应该重新考虑 _data 的类型。

int main
{
    const char data = _data;
    const char *iter = data;
    string strA = pop(iter, 4);
    string strB = pop(iter, 8);
    string strC = pop(iter, 8);
    cout << "strA: '" << strA << "'" << endl;
    cout << "strB: '" << strB << "'" << endl;
    cout << "strC: '" << strC << "'" << endl;
}

制作错误信息

error: no viable conversion from 'string' (aka 'basic_string<char, char_traits<char>, allocator<char> >') to
  'const char'
const char data = _data;