Ok so this might be simple to explain. This works as intended puts item into that slot
 

if (SlotCircuit.getValidItems().contains(handStack.getItem())) {
                    if (slotHandler.getStackInSlot(0).isEmpty()) {
                        slotHandler.insertItem(0, handStack, false);
                    }
                }

 

Below code doesnt put item in slot but deletes item in hand..... why does it not put item into the slot when i put a line of code after it?
 

if (SlotCircuit.getValidItems().contains(handStack.getItem())) {
                    if (slotHandler.getStackInSlot(0).isEmpty()) {
                        slotHandler.insertItem(0, handStack, false);
                        handStack.shrink(1);
                    }
                }

 

Edited September 8, 20205 yr by Mightydanp
add version

We don't know what are and where these came from:

Quote

SlotCircuit

handStack

slotHandler

but apparently, 

handStack.shrink(1);

makes whatever this itemstack came from count - 1

Hand stack is just PlayerEntity.getMainHandStack

 

Slot handler leads to ItemStackHandler

 

If you need more information i will post to full class 

So what i get from this is insert item. Set that into the player which will also break the chain then shrink players hand  ??

So i have to insertItem. Then set the player hand to handstack. Then shrink the handStack 

Have some working example code.

https://github.com/Draco18s/ReasonableRealism/blob/1.14.4/src/main/java/com/draco18s/harderores/block/MillstoneBlock.java#L65-L67

Edited September 8, 20205 yr by Draco18s

ItemStack heldItem = player.getHeldItem(hand);
		if(!heldItem.isEmpty()) {
			MillstoneTileEntity te = (MillstoneTileEntity)world.getTileEntity(pos);
			IItemHandler inventory = te.getCapability(CapabilityItemHandler.ITEM_HANDLER_CAPABILITY, Direction.UP).orElse(null);
			if(inventory == null) return false;
			ItemStack stack = heldItem.copy();
			stack.setCount(1);
			stack = inventory.insertItem(0, stack, true);
			if(stack.isEmpty()) {
				stack = inventory.insertItem(0, heldItem.split(1), false);

Can your code. I dont understand where you remove 1 from handItem

15 minutes ago, Mightydanp said:

heldItem.split(1)

 

Ik not at home but split splits the stack or what  exactly

To the surprise of no one: Yes. It splits the stack such that the returned stack has the indicated size (or the original's size, whichever is lower) and the original stack is reduced by that amount (to a minimum of zero).

 

Oh and this:

27 minutes ago, Mightydanp said:

stack = inventory.insertItem(0, stack, true);

Is what checks to see if the item can be inserted, regardless of what stack is currently in the TE's inventory.

Edited September 8, 20205 yr by Draco18s

Alright what is simulate mean? I looked throw the code but didnt understand exactly what that ment

  You want to do it for real, or just simulate it.

Oh so wheb you do 

stack = inventory.insertItem(0, stack, true);

it takes the item and set it in slot 0 but it doesnt actually move the item?

thats why you can do

 

				stack = inventory.insertItem(0, heldItem.split(1), false)

 

1 minute ago, diesieben07 said:

No it does not "set it in slot 0".

I am confused then so it makes it look like it set that item in slot 0 but doesnt ?

stack = inventory.insertItem(0, stack, true);
			if(stack.isEmpty()) {
				stack = inventory.insertItem(0, heldItem.split(1), false);
Can your code. I dont understand where you remove 1 from handItem

Ok so line 1 will visually put an item in slot 0. Which will make stack empty

This will let it pass the of statement and then you insert the item and split it 

6 minutes ago, Mightydanp said:

Ok so line 1 will visually put an item in slot 0.

No. The first line calculates what would be left over if the stack were to be inserted.

Oh ok so you do that to make sure that that item can go into the slot so that it wont delete the item in your hand if it cant

Yes.

But keep in mind that that is not the only way these methods can be used. That's what it does here because only 1 of whatever item is trying to be inserted at a time. If we wanted to do more than 1, it gets a little trickier, but the general process is still the same:

1. Attempt to insert
2. Check if the remainder has a smaller stack size than what we tried to insert
3. Actually split off and insert only the amount that will fit

 

So i would go get stacksize and whatever minus the size of the stack we just similated

One thing that confuses me that you did handStack.split(1) inside the method but is that going to split the handstack to 0 but that will make it set nothing in slot 1 or will it set that item on the slot then split handStack to 0

Vanilla treats stacks of size zero as identical to air. You don't have to do anything about it.

That's why .isEmpty() is a thing.

I understand that but You put handstack.split(1) in the insertItem. How would that set the item in slot 1 and then split it i am confused i thought it would take the handstack and split it then you would be left with air making it not work as intended

50 minutes ago, Mightydanp said:

I understand that but You put handstack.split(1) in the insertItem. How would that set the item in slot 1 and then split it

There are two functions here.

One is inserting a stack into a slot.

The other is splitting the item stack in the player's hand and returning a new stack.

The former is taking as a parameter the result of the latter.

50 minutes ago, Mightydanp said:

i am confused i thought it would take the handstack and split it then you would be left with air making it not work as intended

handstack is indeed left empty if it was a stack of 1 item.

Empty stacks that are not air, and stacks of air, are functionally identical as far as all of vanilla's code is concerned. The whole point of split() is that you have TWO stacks after calling it, the original (now empty) and the new stack (with one item) that gets inserted into the tile entity. That's how the player's held item is cleared.