Solved More efficient?

Wondering which way would be more efficient or does it really matter?

Code:java

if(item.getType()==Material.MONSTER_EGG && item.getData().getData()==(byte)50)
        {
            p.sendMessage("tom");
            if(!p.hasPermission("craftstuff.monsteregg") || !p.hasPermission("craftstuff.*"))
            {
                p.sendMessage(ChatColor.RED + "You don't have permission to craft this item!");
 
                event.getInventory().setResult(null);
            }
        }
        if(item.getType()==Material.MONSTER_EGG && item.getData().getData()==(byte)51)
        {
            p.sendMessage("tom");
            if(!p.hasPermission("craftstuff.monsteregg") || !p.hasPermission("craftstuff.*"))
            {
                p.sendMessage(ChatColor.RED + "You don't have permission to craft this item!");
 
                event.getInventory().setResult(null);
            }
        }

or

Code:java

if(item.getType()==Material.MONSTER_EGG && item.getData().getData()==(byte)50 || item.getType()==Material.MONSTER_EGG && item.getData().getData()==(byte)51)
        {
            p.sendMessage("tom");
            if(!p.hasPermission("craftstuff.monsteregg") || !p.hasPermission("craftstuff.*"))
            {
                p.sendMessage(ChatColor.RED + "You don't have permission to craft this item!");
 
                event.getInventory().setResult(null);
            }
        }

 

RonaldMcDerp As far as I know they are both identical, it's just the matter of readability.

 

jthort
Cool, thanks!

 

The -most- efficient check would be:

Code:java

if (item.getType() == Material.MONSTER_EGG && (item.getData().getData() == (byte) 50 || item.getData().getData() == (byte) 51)) {...}

per your second solution.

 

1Rogue
Oh nice. So I dont have to keep putting the monster egg in, just 'or' the data values. Thanks!