Question

This is weird. It is a trivial problem:

A std::string with bits with length multiple of 8, the first 8 is: ‘10011100’.

//Convert each 8 bits of encoded string to bytes unsigned char c = 0; for(size_t i = 0; i < encoded.size(); i += 8) {     for(size_t k = 0; k < 8; k++)     {         c <<= k;         if(encoded.at(i + k) == '1') c += 1;          //Debug for first 8 bits         if(i == 0) cout << 'at k = ' << k << ', c = ' << (int)c << endl;     }     outFile.write(reinterpret_cast<char*>(&c), sizeof(char)); } 

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Yields the output:

at k = 0, c = 1 at k = 1, c = 2 at k = 2, c = 8 at k = 3, c = 65 at k = 4, c = 17 at k = 5, c = 33 at k = 6, c = 64 at k = 7, c = 0 

Which doesn’t make sense. Shifting 2 places left and getting 8 from 2 is impossible. The max value it can have should be 111b = 7d and in this case should be 100b = 4d.

Enlight me.

Answer 1

at k = 0, c = 1 at k = 1, c = 2 at k = 2, c = 8 

That is because:

input = 10011100 c = 0  `k=0, b=1` shift by 0 add 1 => `c = 1`, dec = 1 `k=1, b=0` shift by 1 add 0 => `c = 10`, dec = 2 `k=2, b=0` shift by 2 add 0 => `c = 1000`, dec = 8 

b means ‘current bit’. Possibly you don’t want to shift by k, but by 1 ? If you look for a standard C++ solution, you can use std::bitset:

std::bitset<8> bits('10011100'); unsigned char c = bits.to_ulong(); 

For your output into the stream, you can use the put function:

outFile.put(c); 

It avoids casting pointers and also outputs unformatted (settings like the field-width are ignored).