Question

I have a Visual Studio 2008 C++ application where I receive a bit map (not an image). Each bit that is flipped corresponds to a position on a decode map.

typedef unsigned char BYTE;
const unsigned int COL_COUNT = 8;
const unsigned int ROW_COUNT = 4;

static char g_decode_map[ ROW_COUNT ][ COL_COUNT ] = 
{
    { 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h' },
    { 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p' },
    { 'q', 'r', 's', 't', 'u', 'v', 'w', 'x' },
    { 'y', 'z', ',', '.', ' ', ':', '-', '+' }
};

// current implementation
void Decode( const BYTE bitmap[ ROW_COUNT ], 
             const char decode_map[ ROW_COUNT ][ COL_COUNT ], 
             char decoded[ ROW_COUNT * COL_COUNT ] )
{
    int found = 0;
    for( int i = 0; i < ROW_COUNT; ++i )
    {
        for( int j = 0; j < COL_COUNT; ++j )
        {
            if( std::bitset< COL_COUNT >( bitmap[ i ] ).test( j ) )
            {
                decoded[ found++ ] = g_decode_map[ i ][ COL_COUNT - j - 1 ];
            }
        }
    }
}

int main( int argc, char* argv[] )
{
    BYTE bitmap[ ROW_COUNT ] = { 0x01, 0x80, 0x00, 0x00 };
    // expected output { 'h', 'i' } or { 'i', 'h' } order is unimportant

    char decoded[ ROW_COUNT * COL_COUNT + 1 ] = { };
    Decode( bitmap, g_decode_map, decoded );
    printf( "Decoded: %s\r\n", decoded );
    return 0;
}

My current decode implementation works fine, but it strikes me that there may be a more efficient way of doing this. Can anybody suggest a more performant algorithm?

Answer 1

It would be faster to test whether or not each bit is set with bitwise operations as opposed to creating a bitset for each bit that is tested. Try something like this:

for( int i = 0; i < ROW_COUNT; ++i ) {
    for( int j = 0; j < COL_COUNT; ++j ) {
        if(bitmap[i] & (1 << j)) {
            ...

1 << j produces a mask that only has the bit you are wanting to test set. Bitwise anding the mask with your bitmap byte returns true only if that bit is set in bitmap[i]. The result of this conditional should be equivalent to the result of your conditional, and it should be a lot faster.