I am trying to make a button which is a part to teleport you once pressed but when it is pressed it says: attempt to index nil with 'HummanoidRootPart'
I can't find anything on how to fix my problem specifically.
function click(x)
local y = workspace:FindFirstChild(x).HumanoidRootPart.CFrame
local location = script.Parent.Location.Position
y = CFrame.new(location)
print("Teleported")
end
script.Parent.ClickDetector.MouseClick:connect(click)
Thanks in advance
Where it says "workspace:FindFirstChild(x)", workspace can't find the first child called "x". You used x as a parameter to find in the workspace, which doesn't make sense because workspace:FindFirstChild finds objects by name. Workspace can't find the child with the name "x", or workspace returns nil because it tries to find something that is not a string, so it returns "nil" which is "nil.HumanoidRootPart" which is not allowed.
Consider doing this
function click()
local y = workspace:FindFirstChild("nameOfObjectYouWantToFind").HumanoidRootPart.CFrame
local location = script.Parent.Location.Position
y = CFrame.new(location)
print("Teleported")
end
script.Parent.ClickDetector.MouseClick:Connect(click)
Remember to change "nameOfObjectYouWantToFind" to the name of the thing you want to find.
Related
Someone please help me understand this. I have the following code below. I am trying to append index[i]-1 to an empty array. However I am getting this error: "BoundsError: attempt to access 0-element Array{Any,1} at index [1]" :
sample_size_array = [9,5,6,9,2,6,9]
n_minus_1 = []
array_length = length(sample_size_array)
for i in 1:array_length
n_minus_1[i].append(sample_size_array[i] -1)
end
println(n_minus_1)
If Julia does not understand array[0] then why is i starting at 0 and not at 1?
Your code has two problems:
in the first iteration you are trying to access n_minus_1 array at index 1 while this array is still empty (has 0 length) - this throws you an error;
in Julia you do not invoke methods using a . (this symbol is used for different purposes - in this case it is parsed as field access and also would throw an error later)
To solve both those problems use push! function which appends an element at the end of an array. The code could look like this:
sample_size_array = [9,5,6,9,2,6,9]
n_minus_1 = []
array_length = length(sample_size_array)
for i in 1:array_length
push!(n_minus_1, sample_size_array[i]-1)
end
println(n_minus_1)
However in this case the whole operation can be written even simpler as:
n_minus_1 = sample_size_array .- 1
and you do not need any loop (and here you see another use of . in Julia - in this case we use it to signal that we want to subtract 1 from every element of sample_size_array).
I am trying to create an empty map, that will be then populated within a for loop. Not sure how to proceed in Rascal. For testing purpose, I tried:
rascal>map[int, list[int]] x;
ok
Though, when I try to populate "x" using:
rascal>x += (1, [1,2,3])
>>>>>>>;
>>>>>>>;
^ Parse error here
I got a parse error.
To start, it would be best to assign it an initial value. You don't have to do this at the console, but this is required if you declare the variable inside a script. Also, if you are going to use +=, it has to already have an assigned value.
rascal>map[int,list[int]] x = ( );
map[int, list[int]]: ()
Then, when you are adding items into the map, the key and the value are separated by a :, not by a ,, so you want something like this instead:
rascal>x += ( 1 : [1,2,3]);
map[int, list[int]]: (1:[1,2,3])
rascal>x[1];
list[int]: [1,2,3]
An easier way to do this is to use similar notation to the lookup shown just above:
rascal>x[1] = [1,2,3];
map[int, list[int]]: (1:[1,2,3])
Generally, if you are just setting the value for one key, or are assigning keys inside a loop, x[key] = value is better, += is better if you are adding two existing maps together and saving the result into one of them.
I also like this solution sometimes, where you instead of joining maps just update the value of a certain key:
m = ();
for (...whatever...) {
m[key]?[] += [1,2,3];
}
In this code, when the key is not yet present in the map, then it starts with the [] empty list and then concatenates [1,2,3] to it, or if the key is present already, let's say it's already at [1,2,3], then this will create [1,2,3,1,2,3] at the specific key in the map.
IDL beginner here! Let's say I have two procedures, PRO1 and PRO2. If I receive a command line argument in PRO2, how can I give the argument value to a variable in PRO1?
I have previously tried to make an object reference ,'My', to PRO1, but I receive a syntax error on line 6.
PRO PRO2
opts = ob_new('mg_options)
opts.addOption, 'value', 'v'
opts.parseArgs,error_message = errorMsg
My = obj_new('PRO1')
My.A=opts.get('value')
END
For reference, I attempted to follow these instructions for receiving command line arguments: http://michaelgalloy.com/2009/05/11/command-line-options-for-your-idl-program.html
I had something else here, but I think your example above is actually what you want to avoid, yes? I'm not sure how it ends up being all that different, but if you want to make your procedure an object, you'll have to define an actual object (see here) and create methods for it containing your code functionality. Here's something close-ish.
In a file called pro1__define.pro:
function pro1::Init
self.A = 0L
return, 1
end
pro pro1::process, in_val
self.A = in_val
print, self.A
end
pro pro1__define
struct = {pro1, A:0L}
end
Then in pro2 you would do something like
arg = 2
pro1_obj = pro1()
pro1_obj->process, arg
Depending on which version of IDL you are using you may have to modify the initialization line to the obj_new() syntax.
For a homework assignment, we've been instructed to complete a task without introducing any "side-effects". I've looked up "side-effects" on Wikipedia, and though I get that in theory it means "modifies a state or has an observable interaction with calling functions", I'm having trouble figuring out specifics.
For example, would creating a value that holds a non-compile time result be introducing side effects?
Say I had (might not be syntactically perfect):
val myList = (someFunction x y);;
if List.exists ((=) 7) myList then true else false;;
Would this introduce side-effects? I guess maybe I'm confused on what "modifies a state" means in the definition of side-effects.
No; a side-effect refers to e.g. mutating a ref cell with the assignment operator :=, or other things where the value referred to by a name changes over time. In this case, myList is an immutable value that never changes during the program, thus it is effect-free.
See also
http://en.wikipedia.org/wiki/Referential_transparency_(computer_science)
A good way to think about it is "have I changed anything which any later code (including running this same function again later) could ever possibly see other than the value I'm returning?" If so, that's a side effect. If not, then you can know that there isn't one.
So, something like:
let inc_nosf v = v+1
has no side effects because it just returns a new value which is one more than an integer v. So if you run the following code in the ocaml toplevel, you get the corresponding results:
# let x = 5;;
val x : int = 5
# inc_nosf x;;
- : int = 6
# x;;
- : int = 5
As you can see, the value of x didn't change. So, since we didn't save the return value, then nothing really got incremented. Our function itself only modifies the return value, not x itself. So to save it into x, we'd have to do:
# let x = inc_nosf x;;
val x : int = 6
# x;;
- : int = 6
Since the inc_nosf function has no side effects (that is, it only communicates with the outside world using its return value, not by making any other changes).
But something like:
let inc_sf r = r := !r+1
has side effects because it changes the value stored in the reference represented by r. So if you run similar code in the top level, you get this, instead:
# let y = ref 5;;
val y : int ref = {contents = 5}
# inc_sf y;;
- : unit = ()
# y;;
- : int ref = {contents = 6}
So, in this case, even though we still don't save the return value, it got incremented anyway. That means there must have been changes to something other than the return value. In this case, that change was the assignment using := which changed the stored value of the ref.
As a good rule of thumb, in Ocaml, if you avoid using refs, records, classes, strings, arrays, and hash tables, then you will avoid any risk of side effects. Although you can safely use string literals as long as you avoid modifying the string in place using functions like String.set or String.fill. Basically, any function which can modify a data type in place will cause a side effect.
I have a map:
var sessions = map[string] chan int{}
How do I delete sessions[key]? I tried:
sessions[key] = nil,false;
That didn't work.
Update (November 2011):
The special syntax for deleting map entries is removed in Go version 1:
Go 1 will remove the special map assignment and introduce a new built-in function, delete: delete(m, x) will delete the map entry retrieved by the expression m[x]. ...
Go introduced a delete(map, key) function:
package main
func main () {
var sessions = map[string] chan int{};
delete(sessions, "moo");
}
Copied from Go 1 release notes
In the old language, to delete the entry with key k from the map represented by m, one wrote the statement,
m[k] = value, false
This syntax was a peculiar special case, the only two-to-one assignment. It required passing a value (usually ignored) that is evaluated but discarded, plus a boolean that was nearly always the constant false. It did the job but was odd and a point of contention.
In Go 1, that syntax has gone; instead there is a new built-in function, delete. The call
delete(m, k)
will delete the map entry retrieved by the expression m[k]. There is no return value. Deleting a non-existent entry is a no-op.
Updating: Running go fix will convert expressions of the form m[k] = value, false into delete(m, k) when it is clear that the ignored value can be safely discarded from the program and false refers to the predefined boolean constant. The fix tool will flag other uses of the syntax for inspection by the programmer.
From Effective Go:
To delete a map entry, use the delete built-in function, whose arguments are the map and the key to be deleted. It's safe to do this even if the key is already absent from the map.
delete(timeZone, "PDT") // Now on Standard Time
delete(sessions, "anykey")
These days, nothing will crash.
Use make (chan int) instead of nil. The first value has to be the same type that your map holds.
package main
import "fmt"
func main() {
var sessions = map[string] chan int{}
sessions["somekey"] = make(chan int)
fmt.Printf ("%d\n", len(sessions)) // 1
// Remove somekey's value from sessions
delete(sessions, "somekey")
fmt.Printf ("%d\n", len(sessions)) // 0
}
UPDATE: Corrected my answer.