Binary char array into stringstream and pop from the buffer - c++

Question

I have 20byte binary char array. I want to divide into 3 parts: 4byte, 8byte, 8byte. I implemented it like the following. It works but seems I might be able to use buffer stream. I want to know how to use it.
Now
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
}
Expectation
I want to implement like this.
void main()
{
stringstream ss = _data;
strA = ss.pop(4);
strB = ss.pop(8);
strC = ss.pop(8);
}
Update 1
Thank you guys. I'm trying all of answers you gave me one by one. I'm newbie in c++ so it takes time to understand it. The following is Anders K's one.
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
Update 2
It works using string::substr. Thanks Zakir.
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
}
Update 3
I got an error when I try Scheff's way. This is my fault and I think I can solve it. And I think I should reconsider about _data's type.
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;
}
Make Error Message
error: no viable conversion from 'string' (aka 'basic_string<char, char_traits<char>, allocator<char> >') to
'const char'
const char data = _data;

Answer 1

It is not possible to make a new method for std::stringstream. (At least, I would not recommend this.)
Instead, I would suggest to make it a function. The usage would be similar.
#include <bitset>
#include <iostream>
#include <sstream>
#include <string>
using namespace std;
string pop(istream &in, size_t n)
{
string ret;
while (n--) {
unsigned char byte = (unsigned char)in.get();
ret += bitset<8>(byte).to_string();
}
return ret;
}
int main()
{
string data(
"\x11\x22\x33\x44\x55\x66\x77\x88\x99\xaa"
"\xbb\xcc\xdd\xee\xff\xde\xad\xbe\xef\x00", 20);
istringstream in; in.str(data);
string strA = pop(in, 4);
string strB = pop(in, 8);
string strC = pop(in, 8);
cout << "strA: '" << strA << "'" << endl;
cout << "strB: '" << strB << "'" << endl;
cout << "strC: '" << strC << "'" << endl;
return 0;
}
Output:
strA: '00010001001000100011001101000100'
strB: '0101010101100110011101111000100010011001101010101011101111001100'
strC: '1101110111101110111111111101111010101101101111101110111100000000'
Note:
Using a std::istream makes it applicable to any stream derived from std::istream.
There is no error handling in pop(). Thus, the returned result of pop() might be wrong if the passed stream isn't good() afterwards.
Btw. I agree with the comments that a std::stream might be "over-engineered". Thus, here the "light-weight" version:
#include <bitset>
#include <iostream>
#include <string>
using namespace std;
string pop(const char *&iter, size_t n)
{
string ret;
while (n--) {
ret += bitset<8>((unsigned char)*iter++).to_string();
}
return ret;
}
int main()
{
const char data[] =
"\x11\x22\x33\x44\x55\x66\x77\x88\x99\xaa"
"\xbb\xcc\xdd\xee\xff\xde\xad\xbe\xef\x00";
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;
return 0;
}
The output is identical like above.
Note:
The usage of char[] and char* is much more sensitive for out-of-bound access. Thus, it has to be used carefully.
I'm not quite sure whether the (unsigned char) cast is necessary. As I have often seen "funny" effects concerning char, int and sign extension, I guess it cannot hurt. (I feel better with it.)

Answer 2

I can propose you a very simple alternative using string::substr
#include <iostream>
#include <string>
using namespace std;
int main ()
{
string _data="00010001001000100011001101000100\
0101010101100110011101111000100010011001101010101011101111001100\
1101110111101110111111111101111010101101101111101110111100000000";
int lenA = (4*8); //First 4 Bytes
int lenB = (8*8); //Second 8 Bytes
int lenC = (16*8); //Last 16 Bytes
string strA = _data.substr(0, lenA - 1);
string strB = _data.substr(lenA, lenB - 1);
string strC = _data.substr(lenB, lenC - 1);
std::cout << "strA: " << strA << endl;
std::cout << "strB: " << strB << endl;
std::cout << "strC: " << strC << endl;
return 0;
}
This is neat and simple but gets your job done!
Demo here
Output:-
strA: 0001000100100010001100110100010
strB: 010101010110011001110111100010001001100110101010101110111100110
strC: 100110011010101010111011110011001101110111101110111111111101111010101101101111101110111100000000