Converting integer to a bit representation

Question

How can I convert a integer to its bit representation. I want to take an integer and return a vector that has contains 1\'s and 0\'s of the integer\'s bit representation.

Answer 1

It's not too hard to solve with a one-liner, but there is actually a standard-library solution.

#include 
#include 

std::vector< int > get_bits( unsigned long x ) {
    std::string chars( std::bitset< sizeof(long) * CHAR_BIT >( x )
        .to_string< char, std::char_traits, std::allocator >() );
    std::transform( chars.begin(), chars.end(),
        std::bind2nd( std::minus(), '0' ) );
    return std::vector< int >( chars.begin(), chars.end() );
}

C++0x even makes it easier!

#include 

std::vector< int > get_bits( unsigned long x ) {
    std::string chars( std::bitset< sizeof(long) * CHAR_BIT >( x )
        .to_string( char(0), char(1) ) );
    return std::vector< int >( chars.begin(), chars.end() );
}

This is one of the more bizarre corners of the library. Perhaps really what they were driving at was serialization.

cout << bitset< 8 >( x ) << endl; // print 8 low-order bits of x