In MQL4, I have the following line
int OP_TYPE = int(0.5((1+f)*OP_BUY+(1-f)*OP_SELL));
which gives the error: '+' - some operator expected.
What am I supposed to do in this case?
You cannot infer calc types using MQL4, make sure you explicitly type your calculation in full.
int OP_TYPE = int(0.5*((1+f)*OP_BUY+(1-f)*OP_SELL));
Related
According to Rascal's documentation, the "?" operator can be used to query if a variable is "defined".
For example:
int u=1;
int v; // Defined but uninitialised
u = v?2;
v is uninitialised and therefore u will get the value 2.
However, doing this flags a "Warning: deprecated feature: run-time check on variable initialisation"
Hence the question, what is the non-deprecated way to do what the ? operator did in Rascal?
You can check with the IsDefined operator only things that can in principle be "undefined". Variables are not in that class; they were accidentally and now we are deprecating that behavior. In principle, there exists no null or undefined value in Rascal.
Having said that there are situations with maps and keyword fields of nodes and algebraic constructors where it is possible that a declared name does not exist at runtime. So:
myMap[myKey]?def; // a map does not have to have the key
myCons.myKeywordField?def ; // a keyword field does not have to be set
The isDefined operator is part of the assignment syntax on the left-hand side, as explained here: https://www.rascal-mpl.org/docs/Rascal/Statements/Assignment/IsDefined/
Also, the same syntax can be used as an expression: https://www.rascal-mpl.org/docs/Rascal/Expressions/Values/Boolean/IfDefinedElse/
Again, checking variables for undefinedness does not make sense since variables are always defined in Rascal. It is a static error otherwise. Defined but uninitialized variables are for making matching patterns look more elegant:
int i; int j; // here they are declared with a type
// here they are not defined and may not be used
if (<i, j> := <1,2>) { // here they are bound/defined
// here they can be used
}
// here i and j are not defined again and may not be used
Here is a function:
let newPositions : PositionData list =
positions
|> List.filter (fun x ->
let key = (x.Instrument, x.Side)
match brain.Positions.TryGetValue key with
| false, _ ->
// if we don't know the position, it's new
true
| true, p when x.UpdateTime > p.UpdateTime ->
// it's newer than the version we have, it's new
true
| _ ->
false
)
it compiles at expected.
let's focus on two lines:
let key = (x.Instrument, x.Side)
match brain.Positions.TryGetValue key with
brain.Positions is a Map<Instrument * Side, PositionData> type
if I modify the second line to:
match brain.Positions.TryGetValue (x.Instrument, x.Side) with
then the code will not compile, with error:
[FS0001] This expression was expected to have type
'Instrument * Side'
but here has type
'Instrument'
but:
match brain.Positions.TryGetValue ((x.Instrument, x.Side)) with
will compile...
why is that?
This is due to method call syntax.
TryGetValue is not a function, but a method. A very different thing, and a much worse thing in general. And subject to some special syntactic rules.
This method, you see, actually has two parameters, not one. The first parameter is a key, as you expect. And the second parameter is what's known in C# as out parameter - i.e. kind of a second return value. The way it was originally meant to be called in C# is something like this:
Dictionary<int, string> map = ...
string val;
if (map.TryGetValue(42, out val)) { ... }
The "regular" return value of TryGetValue is a boolean signifying whether the key was even found. And the "extra" return value, denoted here out val, is the value corresponding to the key.
This is, of course, extremely awkward, but it did not stop the early .NET libraries from using this pattern very widely. So F# has special syntactic sugar for this pattern: if you pass just one parameter, then the result becomes a tuple consisting of the "actual" return value and the out parameter. Which is what you're matching against in your code.
But of course, F# cannot prevent you from using the method exactly as designed, so you're free to pass two parameters as well - the first one being the key and the second one being a byref cell (which is F# equivalent of out).
And here is where this clashes with the method call syntax. You see, in .NET all methods are uncurried, meaning their arguments are all effectively tupled. So when you call a method, you're passing a tuple.
And this is what happens in this case: as soon as you add parentheses, the compiler interprets that as an attempt to call a .NET method with tupled arguments:
brain.Positions.TryGetValue (x.Instrument, x.Side)
^ ^
first arg |
second arg
And in this case it expects the first argument to be of type Instrument * Side, but you're clearly passing just an Instrument. Which is exactly what the error message tells you: "expected to have type 'Instrument * Side'
but here has type 'Instrument'".
But when you add a second pair of parens, the meaning changes: now the outer parens are interpreted as "method call syntax", and the inner parens are interpreted as "denoting a tuple". So now the compiler interprets the whole thing as just a single argument, and all works as before.
Incidentally, the following will also work:
brain.Positions.TryGetValue <| (x.Instrument, x.Side)
This works because now it's no longer a "method call" syntax, because the parens do not immediately follow the method name.
But a much better solution is, as always, do not use methods, use functions instead!
In this particular example, instead of .TryGetValue, use Map.tryFind. It's the same thing, but in proper function form. Not a method. A function.
brain.Positions |> Map.tryFind (x.Instrument, x.Side)
Q: But why does this confusing method even exist?
Compatibility. As always with awkward and nonsensical things, the answer is: compatibility.
The standard .NET library has this interface System.Collections.Generic.IDictionary, and it's on that interface that the TryGetValue method is defined. And every dictionary-like type, including Map, is generally expected to implement that interface. So here you go.
In future, please consider the Stack Overflow guidelines provided under How to create a Minimal, Reproducible Example. Well, minimal and reproducible the code in your question is, but it shall also be complete...
…Complete – Provide all parts someone else needs to reproduce your
problem in the question itself
That being said, when given the following definitions, your code will compile:
type Instrument() = class end
type Side() = class end
type PositionData = { Instrument : Instrument; Side : Side; }
with member __.UpdateTime = 0
module brain =
let Positions = dict[(Instrument(), Side()), {Instrument = Instrument(); Side = Side()}]
let positions = []
Now, why is that? Technically, it is because of the mechanism described in the F# 4.1 Language Specification under §14.4 Method Application Resolution, 4. c., 2nd bullet point:
If all formal parameters in the suffix are “out” arguments with byref
type, remove the suffix from UnnamedFormalArgs and call it
ImplicitlyReturnedFormalArgs.
This is supported by the signature of the method call in question:
System.Collections.Generic.IDictionary.TryGetValue(key: Instrument * Side, value: byref<PositionData>)
Here, if the second argument is not provided, the compiler does the implicit conversion to a tuple return type as described in §14.4 5. g.
You are obviously familiar with this behaviour, but maybe not with the fact that if you specify two arguments, the compiler will see the second of them as the explicit byref "out" argument, and complains accordingly with its next error message:
Error 2 This expression was expected to have type
PositionData ref
but here has type
Side
This misunderstanding changes the return type of the method call from bool * PositionData to bool, which consequently elicits a third error:
Error 3 This expression was expected to have type
bool
but here has type
'a * 'b
In short, your self-discovered workaround with double parentheses is indeed the way to tell the compiler: No, I am giving you only one argument (a tuple), so that you can implicitly convert the byref "out" argument to a tuple return type.
Consider below struct:
typedef struct _Index {
NSInteger category;
NSInteger item;
} Index;
If I use this struct as a property:
#property (nonatomic, assign) Index aIndex;
When I access it without any initialization right after a view controller alloc init, LLDB print it as:
(lldb) po vc.aIndex
(category = 0, item = 0)
(lldb) po &_aIndex
0x000000014e2bcf70
I am a little confused, the struct already has valid memory address, even before I want to allocate one. Does Objective-C initialize struct automatically? If it is a NSObject, I have to do alloc init to get a valid object, but for C struct, I get a valid struct even before I tried to initialize it.
Could somebody explains, and is it ok like this, not manually initializing it?
To answer the subquestion, why you cannot assign to a structure component returned from a getter:
(As a motivation this is, because I have read this Q several times.)
A. This has nothing to do with Cbjective-C. It is a behavior stated in the C standard. You can check it for simple C code:
NSMakeSize( 1.0, 2.0 ).width = 3.0; // Error
B. No, it is not an improvement of the compiler. If it would be so, a warning would be the result, not an error. A compiler developer does not have the liberty to decide what an error is. (There are some cases, in which they have the liberty, but this are explicitly mentioned.)
C. The reason for this error is quite easy:
An assignment to the expression
NSMakeSize( 1.0, 2.0 ).width
would be legal, if that expression is a l-value. A . operator's result is an l-value, if the structure is an l-value:
A postfix expression followed by the . operator and an identifier designates a member of a structure or union object. The value is that of the named member,82) and is an lvalue if the first expression is an lvalue.
ISO/IEC 9899:TC3, 6.5.2.3
Therefore it would be assignable, if the expression
NSMakeSize( 1.0, 2.0 )
is an l-value. It is not. The reason is a little bit more complex. To understand that you have to know the links between ., -> and &:
In contrast to ., -> always is an l-value.
A postfix expression followed by the -> operator and an identifier designates a member of a structure or union object. The value is that of the named member of the object to which the first expression points, and is an lvalue. 83)
Therefore - that is what footnote 83 explains – ->, &, and . has a link:
If you can calculate the address of a structure S having a component C with the & operator, the expression (&S)->C is equivalent to S.C. This requires that you can calculate the address of S. But you can never do that with a return value, even it is a simple integer …
int f(void)
{
return 1;
}
f()=5; // Error
… or a pointer …
int *f(void)
{
return NULL;
}
f()=NULL; // Error
You always get the same error: It is not assignable. Because it is a r-value. This is obvious, because it is not clear,
a) whether the way the compiler returns a value, esp. whether he does it in address space.
b) when the time the life time of the returned value is over
Going back to the structure that means that the return value is a r-value. Therefore the result of the . operator on that is a r-value. You are not allowed to assign a value to a r-value.
D. The solution
There is a solution to assign to a "returned structure". One might decide, whether it is good or not. Since -> always is an l-value, you can return a pointer to the structure. Dereferencing this pointer with the -> operator has always an l-value as result, so you can assign a value to it:
// obj.aIndex returns a pointer
obj.aIndex->category = 1;
You do not need #public for that. (What really is a bad idea.)
The semantics of the property are to copy the struct, so it doesn't need to be allocated and initialized like an Objective-C object would. It's given its own space like a primitive type is.
You will need to be careful updating it, as this won't work:
obj.aIndex.category = 1;
Instead you will need to do this:
Index index = obj.aIndex;
index.category = 1;
obj.aIndex = index;
This is because the property getter will return a copy of the struct and not a reference to it (the first snippet is like the second snippet, without the last line that assigns the copy back to the object).
So you might be better off making it a first class object, depending on how it will be used.
Have this function defined in your module:
module Data
int inc(x) = x + 1;
Type this in the console:
rascal> import Data;
rascal> import List;
This works:
rascal> inc(1);
int: 2
But this does not:
rascal> list[int] y = [1,2,3];
rascal> mapper(y, inc);
|rascal://<path>|: insert into collection not supported on value and int
☞ Advice
But it works if inc(...)'s parameter type is declared:
int inc(int x) = x + 1;
So why does not having this type declaration work for using the inc(...) function directly, but not for passing that function to mapper(...)?
Because Rascal's type checker is still under development, you are not warned if you make a small mistake like forgetting to provide a type for a function parameter. It may still work, accidentally, in some circumstances but you are guaranteed to run into trouble somewhere as you've observed. The reason is that type inference for function parameters is simply not implemented as a feature. This is a language design decision with the intent of keeping error messages understandable.
So, this is not allowed:
int f(a) = a + 1;
And, it should be written like this:
int f(int a) = a + 1;
I consider it a bug that the interpreter doesn't complain about an untyped parameter. It is caused by the fact that we reuse the pattern matching code for both function parameters and inline patterns. [edit: issue has been registered at https://github.com/cwi-swat/rascal/issues/763]
In your case the example works because dynamically the type of the value is int and addition does not check the parameter types. The broken example breaks because the interpreter does checks the type of the function parameter at the call-site (which defaulted to value for the untyped parameter).
i am a beginner of erlang.
this is my code:
-module(squsum).
-export([main/0]).
ssum(1) -> 1;
ssum(N) -> N*N + ssum(N-1).
main() ->
{_,T} = io:fread("","~d"),
io:fwrite("~p~n",[ssum(T)]).
why will it have such a problem?
** exception error: an error occurred when evaluating an arithmetic expression
in function squsum:ssum/1 (squsum.erl, line 5)
in call from squsum:main/0 (squsum.erl, line 9)
T is a list containing the value, I guess it is the symmetric point of view of IO:fwrite. so simply use:
main() -> {ok,[T]} = io:fread("","~d"), io:fwrite("~p~n",[ssum(T)]).
I'm not too familiar with erlang, but if it is at all possible for T to be negative, then your recursive function would never end.
Similarly, if T=0, the function would never end.
This is really on comment on Pascal's answer: just to add that io:fread/2 returns a list of the terms specified in the format string --- even if the format string has just one term, it returns a list with one element.
The erlang documentation is quite good. io:fread/2 is at http://erlang.org/doc/man/io.html#fread-2