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Hello, StackOverflow community! I am working on this Lua game, and I was testing to see if it would change the text on my TextLabel to the Bitcoins current worth, I was utterly disappointed when nothing showed up.
I have tried to do research on Google, and my code seems to be just right.
Code
Change = false
updated = false
while Change[true] do --While change = true do
worth = math.random(1,4500) --Pick random number
print('Working!') --Say its working
Updated = true --Change the updated local var.
end --Ending while loop
script.Parent.TextLabel.Text.Text = 'Bitcoin is currently worth: ' .. worth
--Going to the Text, and changing in to a New worth.
while Updated[false] do --While updated = false do
wait(180) --Wait
Change = true --After waits 3 minutes it makes an event trigger
end -- Ending while loop
wait(180) --Wait
Updated = false --Reseting Script.
I expect the output on the Label to be a random number.
I can't really speak to roblox, but there are a couple of obvious problems with your code:
Case
You have confusion between capitalized ("Updated", "Change") and lowercase ("updated", "change" [in commented while statement]), which will fail. See, for example:
bj#bj-lt:~$ lua
Lua 5.2.4 Copyright (C) 1994-2015 Lua.org, PUC-Rio
> Updated = true
> print(Updated)
true
> print(updated)
nil
So be super-careful about what identifiers you capitalize. In general, most programmers leave variables like that in all-lowercase (or sometimes things like camelCase). I suppose there might be some oddball lua runtime out there that is case-insensitive, but I don't know of one.
Type misuse.
Updated is a boolean (a true/false value), so the syntax:
while Change[true] do
...is invalid. See:
> if Updated[true] then
>> print("foo")
>> end
stdin:1: attempt to index global 'Updated' (a boolean value)
stack traceback:
stdin:1: in main chunk
[C]: in ?
Note also that the "While change == true do" is also wrong because of case ("While" is not valid lua, but "while" is).
Lastly:
Lack of threading.
You have basically two different things that you're trying to do at once, namely randomly change the "worth" variable as fast as possible (it's in a loop) and see a set a label to match it (it looks like you probably want it to change constantly). This requires two threads of operation (one to change worth and another to read it and stick it on the label). You've written this like you're assuming you have a spreadsheet or something and that. What your code is actually doing is:
Setting some variables
Updating worth indefinitely, printing 'Working!' a bunch, and...
Never stopping
The rest of the code never runs, because the rest of the code isn't in a background thread (basically the first bit monopolizes the runtime and never yields to everything else).
Lastly, even if the top code was running in the background, you only set the Text label one-time to exactly "Bitcoin is currently worth: 3456" (or some similar number) one time. The way this is written there won't be any updates thereafter (and, if it runs once before the other thread has warmed up, it might not be set to anything useful at all).
My guess is that your runtime is spitting out errors left and right due to the identifier problems and/or is running in a tight infinite loop and never actually getting to the label refresh logic.
BJ Black has given an excellent description of the issues with the syntax, so I'll try to cover the Roblox piece of this. In order for this kind of thing to work properly in a Roblox game, here are some assumptions to double check :
Since we are working with a TextLabel, is it inside a ScreenGui? Or a SurfaceGui?
If it's in a ScreenGui, make sure that ScreenGui is in StarterGui, and is this code in a LocalScript
If it's in a SurfaceGui, make sure that SurfaceGui is adorning a Part and this code
is in a Script
After you checked all those pieces, maybe this is closer to what you were thinking :
-- define the variables we're working with
local textLabel = script.Parent.TextLabel
local worth = 0
-- create an infinite loop
spawn(function()
while true do
--Pick random number
worth = math.random(1,4500)
-- update the text of the label with the new worth
textLabel.Text = string.format("Bitcoin is currently worth: %d", worth)
-- wait for 3 minutes, then loop
wait(180)
end
end)
I removed Updated and Changed because all they were doing was deciding whether or not to change the value. The flow of your loop was:
do nothing and display an undefined number. Wait 3 minutes
update the number, display it, wait 6 minutes
repeat 1 and 2.
So hopefully this is a little clearer and closer to what you were thinking.
This question already has answers here:
Lua for loop reduce i? Weird behavior [duplicate]
(3 answers)
Closed 7 years ago.
im trying this in lua:
for i = 1, 10,1 do
print(i)
i = i+2
end
I would expect the following output:
1,4,7,10
However, it seems like i is getting not affected, so it gives me:
1,2,3,4,5,6,7,8,9,10
Can someone tell my a bit about the background concept and what is the right way to modify the counter variable?
As Colonel Thirty Two said, there is no way to modify a loop variable in Lua. Or rather more to the point, the loop counter in Lua is hidden from you. The variable i in your case is merely a copy of the counter's current value. So changing it does nothing; it will be overwritten by the actual hidden counter every time the loop cycles.
When you write a for loop in Lua, it always means exactly what it says. This is good, since it makes it abundantly clear when you're doing looping over a fixed sequence (whether a count or a set of data) and when you're doing something more complicated.
for is for fixed loops; if you want dynamic looping, you must use a while loop. That way, the reader of the code is aware that looping is not fixed; that it's under your control.
When using a Numeric for loop, you can change the increment by the third value, in your example you set it to 1.
To see what I mean:
for i = 1,10,3 do
print(i)
end
However this isn't always a practical solution, because often times you'll only want to modify the loop variable under specific conditions. When you wish to do this, you can use a while loop (or if you want your code to run at least once, a repeat loop):
local i = 1
while i < 10 do
print(i)
i = i + 1
end
Using a while loop you have full control over the condition, and any variables (be they global or upvalues).
All answers / comments so far only suggested while loops; here's two more ways of working around this problem:
If you always have the same step size, which just isn't 1, you can explicitly give the step size as in for i =start,end,stepdo … end, e.g. for i = 1, 10, 3 do … or for i = 10, 1, -1 do …. If you need varying step sizes, that won't work.
A "problem" with while-loops is that you always have to manually increment your counter and forgetting this in a sub-branch easily leads to infinite loops. I've seen the following pattern a few times:
local diff = 0
for i = 1, n do
i = i+diff
if i > n then break end
-- code here
-- and to change i for the next round, do something like
if some_condition then
diff = diff + 1 -- skip 1 forward
end
end
This way, you cannot forget incrementing i, and you still have the adjusted i available in your code. The deltas are also kept in a separate variable, so scanning this for bugs is relatively easy. (i autoincrements so must work, any assignment to i below the loop body's first line is an error, check whether you are/n't assigning diff, check branches, …)
This is my code:
def is_prime(i)
j = 2
while j < i do
if i % j == 0
return false
end
j += 1
end
true
end
i = (600851475143 / 2)
while i >= 0 do
if (600851475143 % i == 0) && (is_prime(i) == true)
largest_prime = i
break
end
i -= 1
end
puts largest_prime
Why is it not returning anything? Is it too large of a calculation going through all the numbers? Is there a simple way of doing it without utilizing the Ruby prime library(defeats the purpose)?
All the solutions I found online were too advanced for me, does anyone have a solution that a beginner would be able to understand?
"premature optimization is (the root of all) evil". :)
Here you go right away for the (1) biggest, (2) prime, factor. How about finding all the factors, prime or not, and then taking the last (biggest) of them that is prime. When we solve that, we can start optimizing it.
A factor a of a number n is such that there exists some b (we assume a <= b to avoid duplication) that a * b = n. But that means that for a <= b it will also be a*a <= a*b == n.
So, for each b = n/2, n/2-1, ... the potential corresponding factor is known automatically as a = n / b, there's no need to test a for divisibility at all ... and perhaps you can figure out which of as don't have to be tested for primality as well.
Lastly, if p is the smallest prime factor of n, then the prime factors of n are p and all the prime factors of n / p. Right?
Now you can complete the task.
update: you can find more discussion and a pseudocode of sorts here. Also, search for "600851475143" here on Stack Overflow.
I'll address not so much the answer, but how YOU can pursue the answer.
The most elegant troubleshooting approach is to use a debugger to get insight as to what is actually happening: How do I debug Ruby scripts?
That said, I rarely use a debugger -- I just stick in puts here and there to see what's going on.
Start with adding puts "testing #{i}" as the first line inside the loop. While the screen I/O will be a million times slower than a silent calculation, it will at least give you confidence that it's doing what you think it's doing, and perhaps some insight into how long the whole problem will take. Or it may reveal an error, such as the counter not changing, incrementing in the wrong direction, overshooting the break conditional, etc. Basic sanity check stuff.
If that doesn't set off a lightbulb, go deeper and puts inside the if statement. No revelations yet? Next puts inside is_prime(), then inside is_prime()'s loop. You get the idea.
Also, there's no reason in the world to start with 600851475143 during development! 17, 51, 100 and 1024 will work just as well. (And don't forget edge cases like 0, 1, 2, -1 and such, just for fun.) These will all complete before your finger is off the enter key -- or demonstrate that your algorithm truly never returns and send you back to the drawing board.
Use these two approaches and I'm sure you'll find your answers in a minute or two. Good luck!
Do you know you can solve this with one line of code in Ruby?
Prime.prime_division(600851475143).flatten.max
=> 6857
I'm maintaining someone else's Lua code, and Lua is not my preferred language. This is probably a complete noob question, but I can't seem to find the answer on Google or SO...
The following code
if !(v.LastHealth == v:Health()) then
local newColor = {}
newColor.r = v.orgColor.r - (v.orgColor.r - curColor.r) --This is the line the error occurs on
newColor.g = v.orgColor.g - (v.orgColor.g * clrPercent)
newColor.b = v.orgColor.b - (v.orgColor.b * clrPercent)
newColor.a = v.orgColor.a - (v.orgColor.a - curColor.a)
v:SetColor( newColor )
produces the error
attempt to perform arithmetic on field 'r' (a table value)
orgColor (maybe, not totally certain- v.orgColor may be an outdated thing) and curColor are tables that have entries (Uh. I think. bla.x is the same as bla[x] in Lua, right?) r, g, b, and a. Apparently I can't do math on things that come from tables? Should I stow all these values in local variables before working with them? That doesn't seem right.
EDIT:
Printing v.orgColor gives table: 0x40390080, which I assume means it exists and is a table. What's odd is, v.orgColor.r gives another table! That sounds like the cause.
As it turns out, v.orgColor was not, as I had presumed, set by the host program, but was set by the same script as the code sample is from. There was an API change that made a function that used to return four RGBA values instead return a table of those same values; the old code set orgColor.r to the table containing those values, causing the error.
Moral of the story, I suppose, is that you should always make sure you know what's setting the variables you're working with.
For example, if I want to read the middle value from magic(5), I can do so like this:
M = magic(5);
value = M(3,3);
to get value == 13. I'd like to be able to do something like one of these:
value = magic(5)(3,3);
value = (magic(5))(3,3);
to dispense with the intermediate variable. However, MATLAB complains about Unbalanced or unexpected parenthesis or bracket on the first parenthesis before the 3.
Is it possible to read values from an array/matrix without first assigning it to a variable?
It actually is possible to do what you want, but you have to use the functional form of the indexing operator. When you perform an indexing operation using (), you are actually making a call to the subsref function. So, even though you can't do this:
value = magic(5)(3, 3);
You can do this:
value = subsref(magic(5), struct('type', '()', 'subs', {{3, 3}}));
Ugly, but possible. ;)
In general, you just have to change the indexing step to a function call so you don't have two sets of parentheses immediately following one another. Another way to do this would be to define your own anonymous function to do the subscripted indexing. For example:
subindex = #(A, r, c) A(r, c); % An anonymous function for 2-D indexing
value = subindex(magic(5), 3, 3); % Use the function to index the matrix
However, when all is said and done the temporary local variable solution is much more readable, and definitely what I would suggest.
There was just good blog post on Loren on the Art of Matlab a couple days ago with a couple gems that might help. In particular, using helper functions like:
paren = #(x, varargin) x(varargin{:});
curly = #(x, varargin) x{varargin{:}};
where paren() can be used like
paren(magic(5), 3, 3);
would return
ans = 16
I would also surmise that this will be faster than gnovice's answer, but I haven't checked (Use the profiler!!!). That being said, you also have to include these function definitions somewhere. I personally have made them independent functions in my path, because they are super useful.
These functions and others are now available in the Functional Programming Constructs add-on which is available through the MATLAB Add-On Explorer or on the File Exchange.
How do you feel about using undocumented features:
>> builtin('_paren', magic(5), 3, 3) %# M(3,3)
ans =
13
or for cell arrays:
>> builtin('_brace', num2cell(magic(5)), 3, 3) %# C{3,3}
ans =
13
Just like magic :)
UPDATE:
Bad news, the above hack doesn't work anymore in R2015b! That's fine, it was undocumented functionality and we cannot rely on it as a supported feature :)
For those wondering where to find this type of thing, look in the folder fullfile(matlabroot,'bin','registry'). There's a bunch of XML files there that list all kinds of goodies. Be warned that calling some of these functions directly can easily crash your MATLAB session.
At least in MATLAB 2013a you can use getfield like:
a=rand(5);
getfield(a,{1,2}) % etc
to get the element at (1,2)
unfortunately syntax like magic(5)(3,3) is not supported by matlab. you need to use temporary intermediate variables. you can free up the memory after use, e.g.
tmp = magic(3);
myVar = tmp(3,3);
clear tmp
Note that if you compare running times with the standard way (asign the result and then access entries), they are exactly the same.
subs=#(M,i,j) M(i,j);
>> for nit=1:10;tic;subs(magic(100),1:10,1:10);tlap(nit)=toc;end;mean(tlap)
ans =
0.0103
>> for nit=1:10,tic;M=magic(100); M(1:10,1:10);tlap(nit)=toc;end;mean(tlap)
ans =
0.0101
To my opinion, the bottom line is : MATLAB does not have pointers, you have to live with it.
It could be more simple if you make a new function:
function [ element ] = getElem( matrix, index1, index2 )
element = matrix(index1, index2);
end
and then use it:
value = getElem(magic(5), 3, 3);
Your initial notation is the most concise way to do this:
M = magic(5); %create
value = M(3,3); % extract useful data
clear M; %free memory
If you are doing this in a loop you can just reassign M every time and ignore the clear statement as well.
To complement Amro's answer, you can use feval instead of builtin. There is no difference, really, unless you try to overload the operator function:
BUILTIN(...) is the same as FEVAL(...) except that it will call the
original built-in version of the function even if an overloaded one
exists (for this to work, you must never overload
BUILTIN).
>> feval('_paren', magic(5), 3, 3) % M(3,3)
ans =
13
>> feval('_brace', num2cell(magic(5)), 3, 3) % C{3,3}
ans =
13
What's interesting is that feval seems to be just a tiny bit quicker than builtin (by ~3.5%), at least in Matlab 2013b, which is weird given that feval needs to check if the function is overloaded, unlike builtin:
>> tic; for i=1:1e6, feval('_paren', magic(5), 3, 3); end; toc;
Elapsed time is 49.904117 seconds.
>> tic; for i=1:1e6, builtin('_paren', magic(5), 3, 3); end; toc;
Elapsed time is 51.485339 seconds.