I am new to coccinelle and trying to find how can I change argument types in function signatures.
I have a function that takes a pointer of type T1 as an argument
int fn(T1* p)
I would like to change the function to take an argument of type T2 as follows
int fn(T2* p)
Can someone point me to an example that changes argument type . I am able to find examples to changer order and number of args.
Thanks in advance.
I recently had to do something similar.
# change_arg_type #
identifier func;
identifier arg;
type T = int;
##
func(...,
- T arg
+ float arg
, ...) { ... }
Something like that should do the trick.
Related
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.
The following code used to work in Dart 2.6 (very simplified example!):
T plus<T extends num>(T a, T b) => a + b;
Now it fails with: A value of type 'num' can't be returned from method 'plus' because it has a return type of 'T'.
Is there a way to abstract over double/int like intended in the example?
I'm assuming you are passing the --no-implicit-casts flag to the analyzer in order to get this error. The code is correct in the current Dart language, but it contains an implicit cast, so if you choose to disable those, the code will not be accepted.
The reason the code does not work is that the static type of a + b is num (both have type T which is only known to extend num, so they are treated as num, and num.operator+ returns num), and the return type is T which may extend num.
So, the code does an implicit cast from num to T if you allow it.
To make the code compile even without implicit casts, you have to turn it into an explicit cast:
T plus<T extends num>(T a, T b) => (a + b) as T;
Is it possible to store a procedure as a property of a derived type? I was thinking of something along the lines of:
module funcs_mod
public :: add
contains
function add(y,z) result (x)
integer,intent(in) :: y,z
integer :: x
x = y + z
end function
end module
module type_A_mod
use funcs_mod
public :: type_A,set_operator
type type_A
procedure(),pointer,nopass :: operator
end type
contains
subroutine set_operator(A,operator)
external :: operator
type(type_A),intent(inout) :: A
A%operator => operator
end subroutine
function operate(A,y,z) result(x)
type(type_A),intent(in) :: A
integer,intent(in) :: y,z
integer :: x
x = A%operator(y,z)
end function
end module
program test
use type_A_mod
use funcs_mod
type(type_A) :: A
call set_operator(A,add)
write(*,*) operate(A,1,2)
end program
But this doesn't successfully compile. Several errors are displayed including:
1) Syntax error in procedure pointer component
and
2) 'operator' at (1) is not a member of the 'type_a' structure
As well as some unsuccessful use statements. Is there a way to do this correctly? Any help is greatly appreciated.
UPDATE:
I've modified procedure,pointer to procedure(),pointer and now the errors are
1) FUNCTION attribute conflicts with SUBROUTINE attribute in 'operator'
and
2) Can't convert UNKNOWN to INTEGER(4)
Both refer to the line x = A%operator(y,z)
As you have discovered, the syntax for declaring a procedure pointer declaration requires procedure([interface]), pointer [, ...] :: .... You chose procedure(), pointer, nopass :: operator.
The consequence of procedure() is that you are not declaring whether operator is a function or a subroutine. There is nothing untoward in this, but more work then remains in convincing the compiler that you are using the references consistently. Your compiler appears to not believe you.
Rather than go into detail of what the compiler thinks you mean, I'll take a different approach.
You reference A%operator for a structure A of type with that component as the result of the function operate. You say clearly in declaring this latter function that its result is an integer.
Now, assuming that you don't want to do exciting things with type/kind conversion to get to that integer result, we'll take that you always intend for A%operator to be a function with integer result. That means you can declare that procedure pointer component to be a function with integer result.
This still leaves you with choices:
type type_A
procedure(integer),pointer,nopass :: operator
end type
being a function with integer result and implicit interface, and
type type_A
procedure(add),pointer,nopass :: operator
end type
being a function with explicit interface matching the function add.
Your ongoing design choices inform your final decision.
As a final note, you aren't using implicit none. This is important when we consider your line
external :: operator
If operator is a function then (by implicit typing rules) it has a (default) real result. So, you want to change to one of the following
integer, external :: operator
or
procedure(integer) :: operator
or
procedure(add) :: operator
To conclude, and echo the comment by Vladimir F, think very carefully about your design. You currently have constraints from the reference of operate (in the function result and its arguments) that look like you really do know that the component will have a specific interface. If you are sure of that, then please do use procedure(add) as the declaration/
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).
Please, tell me how use functions in Maxima?
I tried this
function(g, u):= (print(g(0)), print(u));
function(x^2, 10);
but it doesn't work
I guess you want to evaluate the first argument with a specific value of the second argument. So maybe you want ev(g, x=0) instead of g(0).
(g(0) works only if g is the name of a function or a lambda expression, i.e., an unnamed function.)
Maybe you can explain in more detail what you want to accomplish.
When defining a function F(Y) you want Y to be a x-dependent input variable for the former, right?
You can directly define the a function F of unknown variable Y and then input some function as argument.
(%i_) F(Y) := diff(Y(t),t) + Y(t^2);
F(sin);
(%o_) F(Y):='diff(Y(t),t,1)+Y(t^2)
(%o_) sin(t^2)+cos(t)
and you are done!
If you try to parse F(foo(x)):=something + foo(x) you get
define: in definition of F, found bad argument foo(x)
There is a more specific way I took from maxima: use function as function argument:
Before defining F(foo), you can tell maxima
"by foo I mean foo(x)": depends(foo,x).
Then you can define a function with variable foo, for example
depends(foo,x);
F(foo) := foo(x^2) + foo(x) + diff(foo(x),x) ;
F(sin);