bitboards.c

#include "def.h"
#include <stdio.h>

// Simple function to print any bitboard which is sent to it. Here we revers the ranks as we in our previous
// calculations we took the ranks in reverse order for simplicity in order for the conversions to be easy and now we
// revert them back again to take them to their orignal form where they start from the user side.

void PrintBitBoard(ULL b)
{
    printf("\n");
    for (int rank = 7; rank >= 0; rank--)
    {
        for (int file = 0; file <= 7; file++)
        {
            int sq120 = fr2sq(file, rank);
            int sq64 = sq120to64[sq120];
            if ((1ULL << sq64) & b)
                printf("X");
            else
                printf("-");
        }
        printf("\n");
    }
}

const int BitTable[64] = {63, 30, 3,  32, 25, 41, 22, 33, 15, 50, 42, 13, 11, 53, 19, 34, 61, 29, 2,  51, 21, 43,
                          45, 10, 18, 47, 1,  54, 9,  57, 0,  35, 62, 31, 40, 4,  49, 5,  52, 26, 60, 6,  23, 44,
                          46, 27, 56, 16, 7,  39, 48, 24, 59, 14, 12, 55, 38, 28, 58, 20, 37, 17, 36, 8};

// Return the first index of the set bit int the bit boards and sets it to 0 code taken from chessprogramming.wkik
// and also the bitboard is used only for this function We can also use the simple brute force method but it may be
// slower.
int POP(ULL* bb)
{
    ULL b = *bb ^ (*bb - 1);
    unsigned int fold = (unsigned)((b & 0xffffffff) ^ (b >> 32));
    *bb &= (*bb - 1);
    return BitTable[(fold * 0x783a9b23) >> 26];
}

// Counts the nuber of set bits in a a bitboard.
int COUNT(ULL b)
{
    int r;
    for (r = 0; b; r++, b &= b - 1)
        ;
    return r;
}