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Posted

I'm having a problem where a bunch of subblocks that are essentially different colors of wool are all coming up in the creative tab with the name "Blackstone Cloth". I'm not quite sure how to resolve this matter, as firmly ordering my code to work hasn't helped

 

private final static String[] subNames =

{

"White", "Orange",  "Magenta", "Light Blue", "Yellow", "Lime Green",

"Pink", "Grey", "Light Grey", "Cyan", "Purple", "Blue", "Brown",

"Green", "Red", "Black"

};

 

public void preInit(FMLPreInitializationEvent event) {

                // Stub Method

        GameRegistry.registerBlock(redstoneCloth, "redstoneCloth");

       

        for (int ix = 0; ix < 16; ix++) {

    ItemStack oldRedCloth = new ItemStack(redstoneCloth, 1);

    ItemStack dye = new ItemStack(Item.dyePowder, 1, ix);

    ItemStack redCloth = new ItemStack(redstoneCloth, 1, ix);

   

    GameRegistry.addShapelessRecipe(redCloth, oldRedCloth, dye);

    LanguageRegistry.addName(redCloth, subNames[ix]+"stone Cloth");

    }

        }

Posted

Objects are passed by value (of object reference) in java (unless the argument an expression). It looks like the single reference is getting modified each time through the loop. But, that explanation doesn't make sense. So, ignore me.

 

I probably would have tried something like this first.

LanguageRegistry.addName(new ItemStack(redstoneClock, 1, ix), subNames[ix]+"stone Cloth");

If that didn't work, then I have to dig deeper. Sorry, I cannot be more helpful, not much to go by.

 

[editted for benefit of future readers with similar problem]

Posted
  On 3/25/2014 at 7:25 AM, sequituri said:

Arguments are passed by reference in java (unless the argument an expression). You are passing the same reference every time in your loop, then altering where it points (not passing a new reference each time).

 

Try this for the LanguageRegistry:

LanguageRegistry.addName(new ItemStack(redstoneClock, 1, ix), subNames[ix]+"stone Cloth");

You can do something similar for the recipe.

Sorry... but he is already creating a new instance of the ItemStack in each iteration of the loop, that is not the answer (it would be if he wasn't).

 

@OP: Can you post your Block or Item code? I'm guessing your issue is you forgot to designate an ItemBlockWithMetadata or return appropriately distinct unlocalized names for each stack. Check here for an example of a block with subtypes. Using that your language registry will work just fine as it is.

Posted
  On 3/25/2014 at 7:34 AM, coolAlias said:

  Quote

Arguments are passed by reference in java (unless the argument an expression). You are passing the same reference every time in your loop, then altering where it points (not passing a new reference each time).

 

Try this for the LanguageRegistry:

LanguageRegistry.addName(new ItemStack(redstoneClock, 1, ix), subNames[ix]+"stone Cloth");

You can do something similar for the recipe.

Sorry... but he is already creating a new instance of the ItemStack in each iteration of the loop, that is not the answer (it would be if he wasn't).

 

@OP: Can you post your Block or Item code? I'm guessing your issue is you forgot to designate an ItemBlockWithMetadata or return appropriately distinct unlocalized names for each stack. Check here for an example of a block with subtypes. Using that your language registry will work just fine as it is.

 

You are more likely correct. I am missing something. Sorry.

Posted
  On 3/25/2014 at 7:47 AM, sequituri said:

  Quote

Arguments are passed by reference in java (unless the argument an expression). You are passing the same reference every time in your loop, then altering where it points (not passing a new reference each time).

 

Try this for the LanguageRegistry:

LanguageRegistry.addName(new ItemStack(redstoneClock, 1, ix), subNames[ix]+"stone Cloth");

You can do something similar for the recipe.

Your remark indicates your lack of java knowledge. Please learn java and then correct people.

Why can't you just admit when you are wrong? Look:

for (int i = 0; i < 4; ++i) {
ItemStack stack = new ItemStack(someItem);
// stack is declared and instantiated as a new reference each iteration, it is NOT the same reference
// and this code will work fine
LanguageRegistry.addName(stack, "some name");
}

Yes, that is exactly the same as what you suggested but with two lines instead of one, which is exactly what the OP is already doing. If you are so awesome at Java, then please, explain to me how that is wrong. I'm just a hobbyist, after all, whereas you are a self-proclaimed Java and C++ developer; enlighten me.

Posted
  On 3/25/2014 at 7:25 AM, sequituri said:

Your remark indicates your lack of java knowledge. Please learn java and then correct people.

 

Wait.. Are you actually serious? You do realise that coolAlias is one of the best modders out?

Former developer for DivineRPG, Pixelmon and now the maker of Essence of the Gods

Posted
  On 3/25/2014 at 8:03 AM, sequituri said:

You did not run that code. Prove me wrong by running it, or shut up.

<sigh> I have run this code many times just fine, but I doubt you will take my word for it. It's disappointing that you persist in your obstinance rather than either a) showing why I'm wrong, or b) admitting that you are wrong and gracefully accepting it and perhaps learning from it, as I will do if you demonstrate a fault in my code. There is nothing wrong with being wrong except for continuing to be wrong when you know or should know otherwise.

 

  Quote

Wait.. Are you actually serious? You do realise that coolAlias is one of the best modders out?

Err, I wouldn't say that, but thanks xD There are plenty of things about Java that I have no clue about, but I get by with what I know.

Posted
  On 3/25/2014 at 8:22 AM, sequituri said:

I explained why it was wrong. A reference to a reference is still a reference. The original reference is easily changed even after passing it to a method. If the method stores the reference, it stores where it points. If that reference is later changed by reassignment, if changes the original as well. Simple java. This is the main reason that objects have a .copy() method. It allows a clean address to be used.

Except Java passes by value, not by reference:

public void aMethod() {
ItemStack stack = new ItemStack(Item.diamond);
bMethod(stack);
System.out.println(stack.toString());
// changing stack to null or anything else here does not affect what becomes of the stack sent
// to bMethod, nor does it change what a reference created in bMethod points to
}

public void bMethod(ItemStack stack) {
ItemStack pointer = stack;
stack = new ItemStack(Item.emerald);
System.out.println(stack.toString()); // prints the emerald stack
System.out.println(pointer .toString()); // prints the diamond stack from aMethod
}

If the stack from aMethod and the pointer from bMethod both point at the same memory location, then whatever one does to that data will be reflected in the other; but changing the location that one of those points to does NOT change what the other one points to. Sorry, but you are still wrong.

Posted
  On 3/25/2014 at 9:29 AM, sequituri said:

Only if the argument is an expression or native type - like int, float, etc.

In C++ that would be true, but in Java all variables are passed by value; for non-native types, the value is always a copy of the memory address (maybe not always, but I'm pretty sure that's the case). Unlike C++, you cannot pass by reference in Java, even when passing a memory address. Java is "pass by reference by value", as they like to say ;)

 

EDIT: Here is a link that explains it probably better than I am: http://stackoverflow.com/questions/40480/is-java-pass-by-reference

Posted
  On 3/25/2014 at 9:42 AM, coolAlias said:

  Quote

Only if the argument is an expression or native type - like int, float, etc.

In C++ that would be true, but in Java all variables are passed by value; for non-native types, the value is always a copy of the memory address (maybe not always, but I'm pretty sure that's the case). Unlike C++, you cannot pass by reference in Java, even when passing a memory address. Java is "pass by reference by value", as they like to say ;)

 

EDIT: Here is a link that explains it probably better than I am: http://stackoverflow.com/questions/40480/is-java-pass-by-reference

 

You can test following: Change the stack size of the ItemStack argument -> it is also changed outside of the method:

doSomething() {
    ItemStack stack = new ItemStack(Block.dirt, 1);
    incrStackSize(stack);
    System.out.println(stack.stackSize); // is now 64 here, too!
}
incrStackSize(ItemStack stack) {
    System.out.println(stack.stackSize); // is here 1
    stack.stackSize = 64;
    System.out.println(stack.stackSize); // is now 64
}

 

If you re-assign the "pointer", it will hold the new address of that reference!

doSomething() {
    ItemStack stack = new ItemStack(Block.dirt, 1);
    incrStackSize(stack);
    System.out.println(stack.stackSize); // is still 1!
}
incrStackSize(ItemStack stack) {
    stack = new ItemStack(Block.sand, 64)
    System.out.println(stack.stackSize); // is here 64
}

 

That is true for C++ pointers as well as for Java "pointers"

The only exception is in VB.NET where you can say that a parameter of a method is a direct reference (ByRef) and thus you can actually assign a new reference, but that's another language entirely...

Don't ask for support per PM! They'll get ignored! | If a post helped you, click the "Thank You" button at the top right corner of said post! |

mah twitter

  Quote

This thread makes me sad because people just post copy-paste-ready code when it's obvious that the OP has little to no programming experience. This is not how learning works.

Posted
  On 3/25/2014 at 10:05 AM, SanAndreasP said:

~snip~

^^ That's what I was trying to say, though in C++ you can pass a pointer by reference, in which case you can change the original pointer's address... the source of many a C++ coder's nightmares, to be sure :P

 

Pretty sure anyway, it's been many many years since I've touched C++, so I might be terribly wrong about that statement.

Posted
  On 3/25/2014 at 10:12 AM, coolAlias said:

  Quote

~snip~

^^ That's what I was trying to say, though in C++ you can pass a pointer by reference, in which case you can change the original pointer's address... the source of many a C++ coder's nightmares, to be sure :P

 

Pretty sure anyway, it's been many many years since I've touched C++, so I might be terribly wrong about that statement.

 

Yes, you can use the & as an operator in C++ like method(&pointer);

 

B2T: Can we see your Item class, I'm pretty sure you forgot to override getUnlocalizedName(...)

https://github.com/SanAndreasP/EnderStuffPlus/blob/master/java/sanandreasp/mods/EnderStuffPlus/item/ItemESPPearls.java#L81-L83

 

Also I suggest you don't use the LanguageRegistry anymore, since in 1.7 it's not there anymore and 1.6.4 already supports .lang files in your resource folder

https://github.com/SanAndreasP/EnderStuffPlus/blob/master/resources/assets/enderstuffp/lang/en_US.lang

Don't ask for support per PM! They'll get ignored! | If a post helped you, click the "Thank You" button at the top right corner of said post! |

mah twitter

  Quote

This thread makes me sad because people just post copy-paste-ready code when it's obvious that the OP has little to no programming experience. This is not how learning works.

Posted
  On 3/25/2014 at 7:34 AM, coolAlias said:

  Quote

Arguments are passed by reference in java (unless the argument an expression). You are passing the same reference every time in your loop, then altering where it points (not passing a new reference each time).

 

Try this for the LanguageRegistry:

LanguageRegistry.addName(new ItemStack(redstoneClock, 1, ix), subNames[ix]+"stone Cloth");

You can do something similar for the recipe.

Sorry... but he is already creating a new instance of the ItemStack in each iteration of the loop, that is not the answer (it would be if he wasn't).

 

@OP: Can you post your Block or Item code? I'm guessing your issue is you forgot to designate an ItemBlockWithMetadata or return appropriately distinct unlocalized names for each stack. Check here for an example of a block with subtypes. Using that your language registry will work just fine as it is.

 

package wrink.dappercraft;

import cpw.mods.fml.relauncher.Side;
import cpw.mods.fml.relauncher.SideOnly;
import net.minecraft.block.Block;
import net.minecraft.block.BlockColored;
import net.minecraft.item.ItemBlock;
import net.minecraft.item.ItemDye;
import net.minecraft.item.ItemStack;
import net.minecraft.util.Icon;

public class RedstoneClothItem extends ItemBlock
{
    public RedstoneClothItem(int par1)
    {
        super(par1);
        this.setMaxDamage(0);
        this.setHasSubtypes(true);
    }

    @SideOnly(Side.CLIENT)

    /**
     * Gets an icon index based on an item's damage value
     */
    public Icon getIconFromDamage(int par1)
    {
        return Block.cloth.getIcon(2, BlockColored.getBlockFromDye(par1));
    }

    /**
     * Returns the metadata of the block which this Item (ItemBlock) can place
     */
    public int getMetadata(int par1)
    {
        return par1;
    }

    @Override
    /**
     * Returns the unlocalized name of this item. This version accepts an ItemStack so different stacks can have
     * different names based on their damage or NBT.
     */
    public String getUnlocalizedName(ItemStack par1ItemStack)
    {
        return super.getUnlocalizedName() + "." + ItemDye.dyeColorNames[RedstoneCloth.getBlockFromDye(par1ItemStack.getItemDamage())];
    }
}

 

package wrink.dappercraft;

import java.util.List;
import java.util.Random;

import cpw.mods.fml.relauncher.Side;
import cpw.mods.fml.relauncher.SideOnly;
import net.minecraft.block.material.Material;
import net.minecraft.client.renderer.texture.IconRegister;
import net.minecraft.creativetab.CreativeTabs;
import net.minecraft.item.ItemDye;
import net.minecraft.item.ItemStack;
import net.minecraft.util.Icon;
import net.minecraft.world.World;

public class RedstoneCloth extends ClothBlock
{

private Icon[] iconArray;

public RedstoneCloth(int par1, Material par2Material) {
	super(par1, par2Material);
	setUnlocalizedName("redCloth").setTextureName("dappercraft:redstoneCloth").setCreativeTab(CreativeTabs.tabBlock);
}

@SideOnly(Side.CLIENT)

    /**
     * From the specified side and block metadata retrieves the blocks texture. Args: side, metadata
     */
    public Icon getIcon(int par1, int par2)
    {
        return this.iconArray[par2 % this.iconArray.length];
    }

    /**
     * Determines the damage on the item the block drops. Used in cloth and wood.
     */
    public int damageDropped(int par1)
    {
        return par1;
    }

    /**
     * Takes a dye damage value and returns the block damage value to match
     */
    public static int getBlockFromDye(int par0)
    {
        return ~par0 & 15;
    }

    /**
     * Takes a block damage value and returns the dye damage value to match
     */
    public static int getDyeFromBlock(int par0)
    {
        return ~par0 & 15;
    }

    @SideOnly(Side.CLIENT)

    /**
     * returns a list of blocks with the same ID, but different meta (eg: wood returns 4 blocks)
     */
    public void getSubBlocks(int par1, CreativeTabs par2CreativeTabs, List par3List)
    {
        for (int j = 0; j < 16; ++j)
        {
            par3List.add(new ItemStack(par1, 1, j));
        }
    }

    @SideOnly(Side.CLIENT)

    /**
     * When this method is called, your block should register all the icons it needs with the given IconRegister. This
     * is the only chance you get to register icons.
     */
    public void registerIcons(IconRegister par1IconRegister)
    {
        this.iconArray = new Icon[16];

        for (int i = 0; i < this.iconArray.length; ++i)
        {
            this.iconArray[i] = par1IconRegister.registerIcon(this.getTextureName() + "_" + ItemDye.dyeItemNames[getDyeFromBlock(i)]);
        }
    }
    
    @SideOnly(Side.CLIENT)
    public void randomDisplayTick(World world, int x, int y, int z, Random random)
    {
    	float f1 = (float)x+.5f;
    	float f2 = (float)y+1.1f;
    	float f3 = (float)z+.5f;
    	float f4 = random.nextFloat()*1.4f-.7f;
    	float f5 = (random.nextFloat()*-1.4f+.7f);
    	
    	world.spawnParticle("reddust", (double) f1+f4, f2, f3+f5, 0, 0, 0);
    }
}

 

Posted

Simple, without one you see a block's name something like this, for instance:

 

"tile.blockName.name"

 

In the .lang file, you just need to put that equal to the correct value like:

 

tile.blockName.name=My Real Block Name

 

If all's well and it's in the right place, the name in game should be:

 

"My Real Block Name"

 

Hope that helped! :D

-Mitchellbrine

 

  Quote

Minecraft can do ANYTHING, it's coded in Java and you got the full power of Java behind you when you code. So nothing is impossible.

It may be freaking fucking hard though, but still possible ;)

 

If you create a topic on Modder Support, live by this motto:

  Quote
I don't want your charity, I want your information
Posted

but it's a multiblock. how would I add different names for different metadata?

 

Also, I changed my code and now all the blocks I added have the proper names except for redstone cloth, which is still blackstone cloth. apparently I typed

 

GameRegistry.registerBlock(redstoneCloth, "redstoneCloth");

 

instead of

 

GameRegistry.registerBlock(redstoneCloth, RedstoneClothItem.class);

Posted

You can either a) have an entry in your lang file for each metadata version of the block (based on the unlocalized name that you return), or b) have one entry for the name of your block and override the getItemStackDisplayName method to return a combination of your block's translated name plus the dye color's translated name.

 

Example:

@Override
public String getItemStackDisplayName(ItemStack stack) {
return StatCollector.translateToLocal(getUnlocalizedName()) + ItemDye.dyeItemNames[stack.getItemDamage() % ItemDye.dyeItemNames.length];
}

Posted

public class RedstoneClothItem extends ItemBlock

 

idk if this is related to the names problem,

but shouldnt that be ItemBlockWithMetadata for the subtyping of the items?

 

I was having issues with block subtypes not long ago

and after reading a CoolAlias post elsewhere about IBWMD

a bunch of my problems cleared up

with proper metadata passing between Block form and Item form of the thing

 

maybe this will help, maybe not

but it sounds like you are having Subtype issues (name in particular) and the IBWMD helps take care of meta/subtype issues

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Existing users might receive a direct 100€ discount or a substantial coupon pack for multiple transactions. The beauty of this system is its simplicity and direct impact on your final price. It's designed to be a seamless experience, requiring nothing more than entering the code to unlock significant savings, making your shopping experience more affordable and enjoyable. How To Earn Temu 100€ Coupons As A New Customer? To earn the Temu coupon code 100€ off and the 100 off Temu coupon code as a new customer, the process is incredibly straightforward and rewarding. You simply need to sign up for a new account on the Temu app or website, and our exclusive code ALB496107 will pave the way for your incredible savings. Upon successfully registering as a new user, you become eligible for a host of introductory benefits. When you proceed to checkout for your very first purchase, you'll be prompted to enter a coupon or promo code. This is where you input ALB496107. 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Our special code, ALB496107, ensures that everyone gets to enjoy fantastic deals: ALB496107: A 100€ discount for the first order, making your initial Temu purchase unbelievably affordable. ALB496107: An extra 30% off on any item, giving you an even deeper discount on your chosen products. ALB496107: A free gift for new Temu customers, a delightful welcome gesture to kickstart your shopping journey. ALB496107: Up to 70% discount on any item on the Temu app, providing massive savings on a wide range of products. ALB496107: A free gift with free shipping in the European Nations, such as Germany, France, Italy, Switzerland, etc., combining convenience with extra value. Pros And Cons Of Using Temu Coupon Code 100€ Off This Month Utilizing the Temu coupon 100€ off code and the Temu 100 off coupon this month comes with a host of exciting advantages, alongside a couple of minor considerations. We want to provide you with a balanced view so you can make the most informed decision for your shopping needs. Pros: Instant 100€ savings applied directly to your purchase. Valid for both new and existing customers, ensuring everyone can benefit. Includes a generous 100€ coupon bundle for future multiple uses. Offers additional percentage discounts (e.g., 30% off) for new buyers. Provides free shipping across a wide range of European countries, adding to your overall savings. Cons: The discount might be a "coupon pack" that requires multiple smaller purchases to fully utilize the 100€ value. Some highly popular or already heavily discounted items might have specific exclusions from the additional 30% off. Terms And Conditions Of Using The Temu Coupon 100€ Off In 2025 Understanding the terms and conditions of using the Temu coupon code 100€ off free shipping and latest Temu coupon code 100€ off ensures a smooth and rewarding shopping experience. We've made these terms as simple as possible so you can enjoy your savings without any hidden surprises. Here are the key points to remember: Our coupon code, ALB496107, doesn’t have any expiration date, and readers can use it anytime they want throughout 2025 and beyond. The coupon code is valid for both new and existing users, ensuring everyone can benefit from these amazing discounts. It is applicable to users in European Nations, such as Germany, France, Italy, Switzerland, Spain, and many other countries across Europe. There are no minimum purchase requirements for using our Temu coupon code ALB496107, making it easy to apply to any order, big or small. While our code offers substantial savings, it cannot always be combined with other promotional coupons or specific flash sales, so always check for the best combination of discounts. Final Note: Use The Latest Temu Coupon Code 100€ Off In conclusion, seizing the opportunity to use the Temu coupon code 100€ off is an absolute must for anyone looking to maximize their savings on the Temu platform. We are committed to helping you make the most of your online shopping. Don't miss out on these incredible benefits that the Temu coupon 100€ off provides, whether you're a first-time shopper or a loyal customer. Happy shopping, and enjoy your fantastic savings! FAQs Of Temu 100€ Off Coupon Is ALB496107 the best Temu coupon code for 100€ off? Yes, ALB496107 is currently the best working coupon offering a flat 100€ off across Europe, alongside various other fantastic benefits for both new and existing users.  Can existing users use the Temu 100€ coupon? Absolutely! Our ALB496107 code is valid for both new and returning users, offering an extra 100€ discount and benefits like a coupon bundle for multiple purchases and free shipping.  Does the Temu coupon code expire? No, our specific coupon code, ALB496107, does not have an expiration date. You can use it confidently anytime you wish to shop on Temu. How many times can I use the Temu 100€ off coupon? While the flat 100€ discount typically applies once per user type (new/existing), the code often unlocks a 100€ coupon bundle that can be used across multiple subsequent orders, maximizing your long-term savings. Can I combine the Temu 100€ off coupon with other discounts? Generally, our 100€ off coupon provides significant standalone savings. While it may not always stack with other specific promotional coupons, it often works on top of existing product discounts on Temu.
    • Where did you get the schematic? Source/Link? And do use an own modpack or a pre-configured from curseforge? If yes, which one On a later time, I can make some tests on my own - but I need the schematic and the modpack name
    • man i dont undderstand what you are trying to say. i put the schematic in the schamatic folder and i used the stick from the forgematica mod and i used the loader and itr just crashed
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