二进制字符数组转换为字符串流并从缓冲区弹出

我有 20 字节的二进制字符数组。我想分为 3 个部分：4 字节、8 字节、8 字节。我像下面这样实现它。它可以工作，但似乎我可以使用缓冲流。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
}

期望

我想像这样实现。

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

更新 1

谢谢你们。我正在一一尝试你给我的所有答案。我是 c++ 的新手，所以理解它需要时间。以下是安德斯·

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

它使用字符串：：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

当我尝试谢夫的方式时，我遇到了一个错误。这是我的错，我想我可以解决它。我想我应该重新考虑_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;
}

生成错误消息

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