Saving 8 booleans into 1 byte - c++

Question

I'm coding a program writing down to a file a large bool area (usually around 512*512 bool variables). It would make great use of saving it in a smart way, I'm thinking about saving them 8 by 8, coding these 8 booleans into one byte of the form:
byte = bit0 * boolean0 | ... | bit7 * boolean7
But I'm not sure how to handle this conversion, though I know to write and read a file byte by byte.
I'm using C++, I've no background in CS but this seems close to the topic of serialization though everything I searched on the subject is really unclear to me. Has it already been implemented, or this there a really simple way to implement it? (I meant saving as much CPU time as possible, my program will write and open millions of these files per instance)
Cheers.
Edit:
With the help of Sean (thank you btw!) I managed to get a bit further but it is still not working, a test of the data after saving and reading tells me that it gets corrupted (as in not reconstructed correctly and so not identical to the initial data) somewhere in the writing, reading or both...
My code will probably help.
Here are the writing lines:
typedef char byte;
ofstream ofile("OUTPUT_FILE");
for(int i=0;i<N/8;i++){
byte encoded = 0;
for(int j=0; j<8; j++){
byte bit = (byte)(tab[(i*8+j)/h][(i*8+j)%h]==1);
encoded = (encoded << 1) | bit;
}
ofile << encoded;
}
and the reading lines:
for(int i=0;i<N/8;i++){ //N is the total number of entries I have in my final array
temp = ifile.get(); //reading the byte in a char
for(int j=0; j<8; j++){ // trying to read each bits in there
if(((temp >> j) & 1) ? '1' : '0' ){
tab[(i*8+j)/h][(i*8+j)%h]=1;
}
else{
tab[(i*8+j)/h][(i*8+j)%h]=-1; //my programs manipulates +1 (TRUE) and -1 (FALSE) making most of the operations easier
}
}
}
ifile.close();
Edit2:
I finally managed to do using the bitset<8> objects, much clearer to me than to manipulate bits inside char. I'll probably update my post later with my working code. I'm still concerned with efficiency, is it much quicker to work with than to bitset do you think?

Answer 1

If you don't need to determine the size of you bit array at runtime, you can use std::bitset
http://en.cppreference.com/w/cpp/utility/bitset

Answer 2

You can encode the bools to an int in a loop like this:
bool flags[8];
// populate flags from somewhere
byte encoded=0;
for(int i=0; i<8; i++)
{
byte bit = (byte)flags[i];
encoded = (encoded << 1) | bit;
}
The code uses the fact that casting a bool to a number yields 1 for true and o for false.
Alternatively you can unroll it:
byte encoded = 0;
encoded |= ((byte)flags[0]) << 7;
encoded |= ((byte)flags[1]) << 6;
encoded |= ((byte)flags[2]) << 5;
encoded |= ((byte)flags[3]) << 4;
encoded |= ((byte)flags[4]) << 3;
encoded |= ((byte)flags[5]) << 2;
encoded |= ((byte)flags[6]) << 1;
encoded |= ((byte)flags[7]);
To convert the byte back to an array of flags you can do something like this:
bool flags[8];
byte encoded= /* some value */
for(int i=0, i<8; i++)
{
bool flag=(bool)(encoded & 1);
flags[7-i]=flag;
encoded>>=1;
}