Now I'm making an endless runner where objects are spawned in front on me randomly.
I was told to make a spawnController and globalController object, so I did. Then this code should be put in the controller under step event
if(tick = 32)
{
tick = 0;
instance_create(room_width,room_height,random(spike,groundBlock));
instance_create(room_width,irandom_range(0,room_height-32));
}
tick += 1;
Is there anything wrong with it because i get an error, which is:
In object spawnController, event Step, action 1 at line 4: Wrong number of arguments to function or script.
instance_create(room_width,irandom_range(0,room_height-32));
The error messages in GM can sometimes be a bit unclear.. But in this case it was pretty clear. It goes about this line. And one of the scripts has too few arguments. Either irandom_range or instance_create you forgot an argument.
irandom_range takes two arguments to make a random number, so that is correct.
instance_create however takes 3 arguments: x,y position and the object from which you wish to create an instance. You're simply missing that argument (and the error tells you that). I think that is a typo as you do it correctly in the creation above.
Manual about instance_create
You have a syntax error here:
instance_create(room_width,irandom_range(0,room_height-32);
There's no closing parentheses or a 3rd argument.
One thing that stood out to me is that you used random instead of choose. Im not sure there is a difference in this situation, but choose allows you to list as many arguments you want.
But the other thing as was pointed out, was that your missing the object you want the 4th life to create. You need to specify what object you want it to make.
instance_create(room_width, irandom_range(0,room_height-32), OBJECT);
Related
I've written a prime-generating function generatePrimes (full code here) that takes input bound::Int64 and returns a Vector{Int64} of all primes up to bound. After the function definition, I have the following code:
println("Generating primes...")
println("Last prime: ", generatePrimes(10^7)[end])
println("Primes generated.")
which prints, unexpectedly,
Generating primes...
9999991
Primes generated.
This output misses the "Last prime: " segment of the second print statement. The output does work as expected for smaller inputs; any input at least up to 10^6, but somehow fails for 10^7. I've tried several workarounds for this (e.g. assigning the returned value or converting it to a string before calling it in a print statement, combining the print statements, et cetera) and discovered some other weird behaviour: if the "Last prime", is removed from the second print statement, for input 10^7, the last prime doesn't print at all and all I get is a blank line between the first and third print statements. These issues are probably related, and I can't seem to find anything online about why some print statements wouldn't work in Julia.
Thanks so much for any clarification!
Edit: Per DNF's suggestion, following are some reductions to this issue:
Removing the first and last print statements doesn't change anything -- a blank line is always printed in the case I outlined and each of the cases below.
println(generatePrimes(10^7)[end]) # output: empty line
Calling the function and storing the last index in a variable before calling println doesn't change anything either; the cases below work exactly the same either way.
lastPrime::Int = generatePrimes(10^7)[end]
println(lastPrime) # output: empty line
If I call the function in whatever form immediately before a println, an empty line is printed regardless of what's inside the println.
lastPrime::Int = generatePrimes(10^7)[end]
println("This doesn't print") # output: empty line
println("This does print") # output: This does print
If I call the function (or print the pre-generated-and-stored function result) inside a println, anything before the function call (that's also inside the println) isn't printed. The 9999991 and anything else there may be after the function call is printed only if there is something else inside the println before the function call.
# Example 1
println(generatePrimes(10^7)[end]) # output: empty line
# Example 2
println("This first part doesn't print", generatePrimes(10^7)[end]) # output: 9999991
# Example 3
println("This first part doesn't print", generatePrimes(10^7)[end], " prints") # output: 9999991 prints
# Example 4
println(generatePrimes(10^7)[end], "prime doesn't print") # output: prime doesn't print
I could probably list twenty different variations of this same thing, but that probably wouldn't make things any clearer. In every single case version of this issue I've seen so far, the issue only manifests if there's that function call somewhere; println prints large integers just fine. That said, please let me know if anyone feels like they need more info. Thanks so much!
Most likely you are running this code from Atom Juno which recently has some issues with buffering standard output (already reported by others and I also sometimes have this problem).
One thing you can try to do is to flush your standard output
flush(stdout)
Like with any unstable bug restarting Atom Juno also seems to help.
I had the same issue. For me, changing the terminal renderer (File -> Settings -> Packages -> julia-client -> Terminal Options) from webgl to canvas (see pic below) seems to solve the issue.
change terminal renderer
I've also encountered this problem many times. (First time, it was triggered after using the debugger. It is probably unrelated but I have been using Julia+Juno for 2 weeks prior to this issue.)
In my case, the code before the println statement needed to have multiple dictionary assignation (with new keys) in order to trigger the behavior.
I also confirmed that the same code ran in Command Prompt (with same Julia interpreter) prints fine. Any hints about how to further investigate this will be appreciated.
I temporarily solve this issue by printing to stderr, thinking that this stream has more stringent flush mechanism: println(stderr, "hello!")
warning: I'm pretty new to R, sniny and co ==> I don't realize whether this question is interesting.
update : It turns out that it is a shiny question and it seems to be a frequent problem, that is not obvious. Please read all answers, they don't address the same cases.
I have a Data base in DB. Is there a difference between DBtoto <- reactive({DB()}) and DBtoto <- reactive({DB}) ? If so, what is it ?
In fact I don't see what BD() (with parentheses) means.
Yes, there's a difference. DB() is a call to the function named DB. DB is the function itself. If it's not a function, then DB() doesn't make sense, and will trigger a run-time error (unless there's another object somewhere which is a function).
reactive() is a Shiny function, that says the value of its argument may change over time. Usually it would make more sense to think the value of the function call would change, but it's (remotely) possible that the function itself could change.
I also found the first answer of What is “object of type ‘closure’ is not subsettable” error in Shiny? addresses this question. To summary, everything created with 'reactive()' in shiny must be referred as a function.
In my example, if DB was reactive (for instance DB <- reactive(read_DataBase())), then DB() must be referred with parenthesis. For instance, to get the attribute 'x', you must write BD()$x. In my 'DBtoto' example above the first expression holds in the case DB is itself reactive.
I was going through swirl() again as a refresher, and I've noticed that the author of swirl says the command ?matrix is the correct form to calling for a help screen. But, when I run ?matrix(), it still works? Is there a difference between having and not having a pair of parenthesis?
It's not specific to the swirl environment (about which I was entirely unaware until 5 minutes ago) That is standard for R. The help page for the ? shortcut says:
Arguments
topic
Usually, a name or character string specifying the topic for which help is sought.
Alternatively, a function call to ask for documentation on a corresponding S4 method: see the section on S4 method documentation. The calls pkg::topic and pkg:::topic are treated specially, and look for help on topic in package pkg.
It something like the second option that is being invoked with the command:
?matrix()
Since ?? is actually a different shortcut one needs to use this code to bring up that page, just as one needs to use quoted strings for help with for, if, next or any of the other reserved words in R:
?'?' # See ?Reserved
This is not based on a "fuzzy logic" search in hte help system. Using help instead of ? gets a different response:
> help("str()")
No documentation for ‘str()’ in specified packages and libraries:
you could try ‘??str()’
You can see the full code for the ? function by typing ? at the command line, but I am just showing how it starts the language level processing of the expressions given to it:
`?`
function (e1, e2)
{
if (missing(e2)) {
type <- NULL
topicExpr <- substitute(e1)
}
#further output omitted
By running matrix and in general any_function you get the source code of it.
I am using custom LLVM pass where if I encounter a store to
where the compiler converts the value to a Constant; e.g. there is an explicit store:
X[gidx] = 10;
Then LLVM will generate this error:
aoc: ../../../Instructions.cpp:1056: void llvm::StoreInst::AssertOK(): Assertion `getOperand(0)->getType() == cast<PointerType>(getOperand(1)->getType())->getElementType() && "Ptr must be a pointer to Val type!"' failed.
The inheritance order goes as: Value<-User<-Constant, so this shouldn't be an issue, but it is. Using an a cast on the ConstantInt or ConstantFP has no effect on this error.
So I've tried this bloated solution:
Value *new_value;
if(isa<ConstantInt>(old_value) || isa<ConstantFP>(old_value)){
Instruction *allocInst = builder.CreateAlloca(old_value->getType());
builder.CreateStore(old_value, allocInst);
new_value = builder.CreateLoad(allocResultInst);
}
However this solution creates its own register errors when different type are involved, so I'd like to avoid it.
Does anyone know how to convert a Constant to a Value? It must be a simple issue that I'm not seeing. I'm developing on Ubuntu 12.04, LLVM 3, AMD gpu, OpenCL kernels.
Thanks ahead of time.
EDIT:
The original code that produces the first error listed is simply:
builder.CreateStore(old_value, store_addr);
EDIT2:
This old_value is declared as
Value *old_value = current_instruction->getOperand(0);
So I'm grabbing the value to be stored, in this case "10" from the first code line.
You didn't provide the code that caused this first assertion, but its wording is pretty clear: you are trying to create a store where the value operand and the pointer operand do not agree on their types. It would be useful for the question if you'd provide the code that generated that error.
Your second, so-called "bloated" solution, is the correct way to store old_value into the stack and then load it again. You write:
However this solution creates its own register errors when different type are involved
These "register errors" are the real issue you should be addressing.
In any case, the whole premise of "converting a constant to a value" is flawed - as you have correctly observed, all constants are values. There's no point storing a value into the stack with the sole purpose of loading it again, and indeed the standard LLVM pass "mem2reg" will completely remove such a sequence, replacing all uses of the load with the original value.
I try to evaluate a field in my report but it fails every time :
= IIf(Fields!lectcrs_hrs.IsMissing,
Round(Fields!lectcrs_fee.Value * "1.00", 2),
Round(Fields!lectcrs_fee.Value * Fields!lectcrs_hrs.Value, 2))
in the case of Fields!lectcrs_hrs.IsMissing = true my field is empty and i find that the reason that the second case Round(Fields!lectcrs_fee.Value * Fields!lectcrs_hrs.Value, 2) contains a missing field Fields!lectcrs_hrs .why it checks the second case if it passes the first one !
How to fix this problem ?
The behavior you are looking for is called "short-circuiting" and, unfortunately, the IIf function in Visual Basic does not offer that. The reason being is that IIf() is a ternary function and, as such, all arguments passed into it are evaluated before the function call occurs. A ternary operator (If() in VB 9+), on the other hand, does support conditional evaluation. However, I do not believe that the If() operator can be used as a part of an expression in SSRS.
Given the fact that you are trying to use a field which may or may not exist at run time, I suggest that you create a custom function in your report to handle the proper checking and return values. For reference, take a look at this blog post, which covers the same scenario.