Difference in bitset<10> and bitset<2>(input[i]) , need explanation - c++

Question

I just learned some simple encryption today and wrote a simple program to convert my text to 10-bit binary. Im not sure if i'm doing it correctly, but the commented section of the code and the actual code has 2 different 10-bit outputs. I am confused. Can someone explain it to me in layman terms?
#include <iostream>
#include <string>
#include <bitset>
#include "md5.h"
using namespace std;
using std::cout;
using std::endl;
int main(int argc, char *argv[])
{
string input ="";
cout << "Please enter a string:\n>";
getline(cin, input);
cout << "You entered: " << input << endl;
cout << "md5 of " << input << ": " << md5("input") << endl;
cout << "Binary is: ";
// cout << bitset<10>(input[1]);
for (int i=0; i<5; i++)
cout << bitset<2>(input[i]);
cout << endl;
return 0;
}

Answer 1

tl;dr : A char is 8 bits, and the string operator[] returns the different chars, as such you accessed different chars and took the first two bits of those. The solution comes in treating a char as exactly that: 8 bits. By doing some clever bit manipulation, we can achieve the desired effect.
The problem
While I still have not completely understood, what you were trying to do, I can answer what a problem could be with this code:
By calling
cout<<bitset<10>(input[1]);
you are reading the 10 bits starting from the second character ( input[0] would start from the first character).
Now, the loop does something entirely different:
for (int i=0; i<5; i++)
cout << bitset<2>(input[i]);
It uses the i-th character of the string and constructs a bitset from it.
The reference of the bitset constructor tells us this means the char is converted to an unsigned long long, which is then converted to a bitset.
Okay, so let's see how that works with a simple input string like
std::string input = "aaaaa";
The first character of this string is 'a', which gives you the 8 bits of '01100001' (ASCII table), and thus the 10 bit bitset that is constructed from that turns out to print
0001100001
where we see a clear padding for the bits to the left (more significant).
On the other hand, if you go through the characters with your loop, you access each character and take only 2 of the bits.
In our case of the character 'a'='01100001', these bits are '01'. So then your program would output 01 five times.
Now, the way to fix it is to actually think more about the bits you are actually accessing.
A possible solution
Do you want to get the first ten bits of the character string in any case?
In that case, you'd want to write something like:
std::bitset<10>(input[0]);
//Will pad the first two bits of the bitset as '0'
or
for(int i=0;i<5;++i){
char referenced = input[i/4];
std::bitset<2>((referenced>>(6-(i%4)*2)));
}
The loop code was redesigned to read the whole string sequentially into 2 bit bitsets.
So since in a char there are 8 bits, we can read 4 of those sets out of a single char -> that is the reason for the "referenced".
The bitshift in the lower part of the loop makes it so it starts with a shift of 6, then 4, then 2, then 0, and then resets to 6 for the next char, etc...
(That way, we can extract the 2 relevant bits out of each 8bit char)
This type of loop will actually read through all parts of your string and do the correct constructions.
A last remark
To construct a bitset directly from your string, you would have to use the raw memory in bits and from that construct the bitset.
You could construct 8 bit bitsets from each char and append those to each other, or create a string from each 8 bit bitset, concatenate those and then use the final string of 1 and 0 to construct a large bitset of arbitrary size.
I hope it helped.