Get the block that is "x" away from the player (in his direction)

Discussion in 'Plugin Development' started by orange451, Aug 1, 2012.

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     public static double lengthdir_x(double len, double dir) {
    return len * Math.cos(Math.toRadians(dir));
    public static double lengthdir_y(double len, double dir) {
    return -len * Math.sin(Math.toRadians(dir));
    public Block getBlockXAway(Player p, double distance) {
    Location ploc = p.getLocation().add(0,2,0);
    double pitch = -ploc.getPitch();
    double direction = -ploc.getYaw()-90;
    int xx = (int) (ploc.getBlockX() + ( lengthdir_x(distance, direction) * lengthdir_x(1, pitch)));
    int yy = (int) (ploc.getBlockY() + (-lengthdir_y(distance, pitch)));
    int zz = (int) (ploc.getBlockZ() + ( lengthdir_y(distance, direction) * lengthdir_x(1, pitch)));
    Location nloc = new Location(ploc.getWorld(), xx, yy, zz);
    Block b = nloc.getBlock();
    return b;
    What I wrote above works perfectly. It finds the player, his direction, and gets the block that is "x" away from the player and still is in his line of sight. However, I want to know whether or not this is the fastest way to go about this. Is it?
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    You're not using any loops, so the difference between your method and any other method will be pretty negligible.
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