sly | JavaScript library for one-directional scrolling | Menu library

Actually, it has a simple answer: infinite loops

Just add a while True (or condition with a counter) in your code.

Example:

You are using the same identifier card for the list items in the nested loops, perhaps you meant it as:

I had this same problem and I got an answer here.

It's is because the line you read is always followed by a \n character. You need to remove it. Hence, just replace that last part of your code with this. .strip() will do for you. str(current_location).strip("\n")

Whenever you read a line from a text file, it adds a \n character to tell that new line started from there. You need to remove that while printing or it will mess up with your current statement

For the first attempt, I'm unsure where the ${PillButtons.buttons} is coming from, maybe you also have a String getButtons() that returns that JSON.

For the second one, you are using the right method in ${PillButtons.getButtonsListObject} (you could also use ${PillButtons.buttonsListObject} as HTL is smart enough to look for the getter) but you also need to print out

In your NoobPy constructor, you print out the result of evaluating the syntax tree (unless it's None, which will happen if you evaluate an assignment):

You are trying to copy the element, instead of it's Value, you have to get it's Value using element.value, also if the value of field variable is null, this means the Element you are trying to get doesn't exist with that Element ID

And also, you were trying to copy the whole element instead of it's value, So Replace the first line in your function with this one:

NOTE: This is NOT an answer.

Pretty much doing the exact same thing, but in vtk.py instead, plus including all other system dependencies, using os.add_dll_directory() does not work:

You can check the below lines of code

Although we call this technique bottom-up parsing, it's really still top-down in the sense that the grammar is applied starting with what will be the root of the parse, which is the start symbol, on this case statement.

What makes the parse "bottom-up" is the construction of the parse tree. Unlike top-down parsing, in which the next tree node is immediately predicted (and added to the parse tree) before the child nodes are parsed, the bottom-up parser first parses the child nodes and only once all the children have been reduced does it add the parent node. Thus, the root node is the last thing added to the tree; the tree has been built from the bottom up. [Note 1]

This gives the parser a lot more flexibility because it doesn't need to know for certain which production to predict. It can predict a number of possibilities and explore them on parallel. (This can be done in linear time using a state machine.)

In effect, condition only becomes relevant when it might be the next thing to parse. In your grammar, that only happens after an IF token, so it doesn't apply to your sample input.

Here's a quick summary of the parsing process. (If you haven't been exposed to finite state machines, this might not be so easy to follow.) It's useful to know that:

• An item is a single production from the grammar with a position marker (written as •) which indicates the current input position.
• An itemset is a set of items. The parser has a finite state machine whose states are itemsets. Since there are a finite number of productions, there are a finite number of possible items and thus a finite number of itemsets. That makes it possible to precompute the finite state machine.
• If an item has the position mark at the end, then the item can be reduced; the fact that the position mark is at the end of the item indicates that the entire production has been recognised. Reducing a production is what actually creates a parse-tree node (or moral equivalent).
• Since this parser has lookahead (of one symbol), each reducible item also has a lookahead set, which is just a set of terminals. The reduction can only be performed if the next input symbol is in this set. I'm not going to get into how this set is computed. This set is usually written after the item surrounded in [...].
• If the position mark is anywhere else in the item, either a shift or goto transition is possible. A shift is performed when a position mark in the current state is just before a terminal and that terminal is the next input token. A goto transition is performed when a position mark is just before a non-terminal and that non-terminal has just been reduced. The goto transition is taken from the state which was active when recognition of the non-terminal started. (The parser uses stack to keep track of state history.)
• Both shifts and gotos are handled by selecting the items from the itemset in which the position mark precedes the terminal or non-terminal which has been recognised, and advancing the position mark in those items. The selected items are the itemset for the next state (but see the following point).
• If the position mark is at the beginning of an item which starts with a non-terminal, all of the productions for that non-terminal are added to the itemset, also with their position marks at the beginning. (This may need to be done repetitively.)
• The initial state for the state machine is an item representing the grammar's start symbol followed by an end-of-input mark (usually written as $), with the position mark at the beginning. This itemset is expanded as per the step above.

With your grammar, the state machine's starting state is the itemset →•: