Consider the following Dart code:
class Vec2 {
num x, y;
Vec2(this.x, this.y);
Vec2 operator*(num rhs) => Vec2(x * rhs, y * rhs);
String toString() => "<$x, $y>";
}
void main() => print(Vec2(1, 2) * 3);
The output is as expected:
<3, 6>
However, this only works when the left-hand side of the expression is a Vec2 and the right-hand side is a num. In this case, I want the multiplication operator to be commutative, so I write the following extension:
extension Vec2Util on num {
Vec2 operator*(Vec2 rhs) => Vec2(rhs.x * this, rhs.y * this);
}
One might naturally expect the following code to produce identical output to the first snippet:
void main() {
num x = 3;
print("${x * Vec2(1, 2)}");
}
However, the compiler is instead reporting that The argument type 'Vec2' can't be assigned to the parameter type 'num'. It looks as though the compiler is resolving the multiplication to num operator*(num rhs) in this case and then complaining that my Vec2 can't be passed in as a num operand. Why does the compiler apparently disregard my extension? What is the correct way to create custom commutative operators like this?
You cannot do what you want.
Dart user-definable binary operators are all methods on the first operand. Doing e1 + e2 is kind-of like doing e1.+(e2) where + is the name of the method, except you can't normally call a method +.
In order to be able to do 3 * vector, you need the method to be on 3.
You can't add methods to other people's classes, not without fiddling with their source code, and int is a system class so even that is not possible.
You cannot use extension methods because extension methods do not apply when the receiver type already has an instance method with the same name.
And int defines all the operators.
(It's like, and definitely not coincidentally, that the user-definable operators were chosen from exactly what int needs. That's not new in Dart, the operators go back to C and Java.)
So, you can define an extension method on int, but you can't call it anyway, not without an override:
extension MyInt on int {
Vector operator *(Vector other) => other.scalerMultiplication(this);
}
... MyInt(3) * vector ...
That's more complication than just swapping the operands.
I'm writing a binding for a C library with the help of rust-bindgen for which the function signatures are generated automatically into a bindings.rs as:
#[repr(C)]
struct A {
//...
}
struct B {
//...
}
extern "C" {
pub fn foo(x: *mut A, y: *mut B);
//...
}
I'm not very happy with this signature of foo because I know that x is a pointer to a constant struct. Moreover, I want to apply this idea to improve this signature into something like
extern "C" {
pub fn foo(x: &'_ A, y: &'_ mut B);
}
But binding.rs has a bunch functions like foo and rewriting them by hand is a very time consuming task and I think that macros (or something else) should help. For example, there might exist one (or several) magic macro rewrite!
// hide
mod ffi {
include!("binding.rs"); // so bunch of functions: foo, bar
}
// re-exports
extern "C" {
rewrite!(foo); // should expand to: pub fn foo(x: &'_A, y: &'_ mut B)
rewrite!(bar);
}
I'm at a very early stage of this work. I don't even know if such a problem can be solved by a macro or anything else, so I'm looking for any entry point.
I've cross-posted this question to the Rust user forum.
A declarative macro can't accomplish this but a procedural macro might be able to. With proc_macro2, you can modify the token stream of a function declaration by placing your rewrite attribute on it, e.g.
extern "C" {
#[rustify]
pub fn foo(x: *mut A, y: *mut B);
}
And your rustify macro would substitute *mut Typename with Option<&mut Typename>.
I don't know how you'd change the mut borrow offhand without replacing the original declaration with *const.
typedef (void (^blockType)());
I need cast blocks with different argument types into a same type blockType, and invoke it as the original type later. But there is a issue while casting block type.
The following code works well with any argument type, ...
((blockType)^(BOOL b) {
NSLog(#"BOOL: %d", b);
})(YES); // >> BOOL: 1
((blockType)^(int i) {
NSLog(#"int: %d", i);
})(1); // >> int: 1
((blockType)^(double f) {
NSLog(#"double: %f", f);
})(1.0 / 3); // >> double: 0.333333
((blockType)^(NSString *s) {
NSLog(#"NSString *: %#", #"string");
})(1.0 / 3); // >> NSString *: string
except float:
((blockType)^(float f) {
NSLog(#"float: %f", f);
})(1.0f); // >> float: 0.000000
((blockType)^(float f) {
NSLog(#"float: %f", f);
})(1.0f / 3); // >> float: 36893488147419103232.000000
but it is ok without casting:
(^(float f) {
NSLog(#"float without casting: %f", f);
})(1.0 / 3); // >> float without casting: 0.333333
how to explain and resolve it?
It appears to be a taint of the good old C language. Consider the following code (we can say it is kind of 'translation' of your Obj-C block with issues to C as far as blocks are related to function pointers (see here)):
void test()
{
void (*pEmpty)();
pEmpty = functionFloat;
pEmpty(1.0f / 3);
}
void functionFloat(float f)
{
printf("float: %f", f);
}
If you call test you will see the same result as when you invoke your 'sick' block. Compiler will provide just a warning about incompatible pointers and will let you run. But if you change
void (*pEmpty)();
to
void (*pEmpty)(void);
there will be a compile-time error. Same will happen if you add void explicitly to your void-blocks, e.g. (void (^)(void) instead of (void (^)().
The reason for such behavior explained in the C Standard:
The empty list in a function declarator that is not part of a
definition of that function specifies that no information about the number or types of the
parameters is supplied.§6.7.6.3-14 (p.134)
Thus, as it doesn't mean that there are no parameters but rather no info about them, cast passes fine.
The problem with unexpected output is the following:
A pointer to a function of one type may be converted to a pointer to a function of another
type and back again; the result shall compare equal to the original pointer. If a converted
pointer is used to call a function whose type is not compatible with the referenced type,
the behavior is undefined.§6.3.2.3-8 (p.56)
and
If the function is defined with a type that is not compatible with the type (of the
expression) pointed to by the expression that denotes the called function, the behavior is
undefined.§6.5.2.2-9 (p.82)
So, it seems that the solution here is just like #jtbandes said: don't mess block types and re-design this part of code to avoid such casts.
Explain: Calling the block as blockType- (void (^)()), the block is treated as (void (^)(double)).
Resolve: Must cast the block back to (void (^)(float)) when invoking.
Does Rascal support function pointers or something like this to do this like Java Interfaces?
Essentially I want to extract specific (changing) logic from a common logic block as separate functions. The to be used function is passed to the common block, which then call this function. In C we can do this with function pointers or with Interfaces in Java.
First I want to know how this general concept is called in the language design world.
I checked the Rascal Function Helppage, but this provide no clarification on this aspect.
So e.g. I have:
int getValue(str input) {
.... }
int getValue2(str input){
... }
Now I want to say:
WhatDatatype? func = getValue2; // how to do this?
Now I can pass this to an another function and then:
int val = invoke_function(func,"Hello"); // how to invoke?, and pass parameters and get ret value
Tx,
Jos
This page in the tutor has an example of using higher-order functions, which are the Rascal feature closest to function pointers:
http://tutor.rascal-mpl.org/Rascal/Rascal.html#/Rascal/Concepts/Functions/Functions.html
You can define anonymous (unnamed) functions, called closures in Java; assign them to variables; pass them as arguments to functions (higher-order functions); etc. Here is an example:
rascal>myfun = int(int x) { return x + 1; };
int (int): int (int);
rascal>myfun;
int (int): int (int);
rascal>myfun(3);
int: 4
rascal>int applyIntFun(int(int) f, int x) { return f(x); }
int (int (int), int): int applyIntFun(int (int), int);
rascal>applyIntFun(myfun,10);
int: 11
The first command defines an increment function, int(int x) { return x + 1; }, and assigns this to variable myfun. The rest of the code would work the same if instead this was
int myfun(int x) { return x + 1; }
The second command just shows the type, which is a function that takes and returns int. The third command calls the function with value 3, returning 4. The fourth command then shows a function which takes a function as a parameter. This function parameter, f, will then be called with argument x. The final command just shows an example of using it.
I wonder what this means in F#.
“a function taking an integer, which returns a function which takes an integer and returns an integer.”
But I don't understand this well.
Can anyone explain this so clear ?
[Update]:
> let f1 x y = x+y ;;
val f1 : int -> int -> int
What this mean ?
F# types
Let's begin from the beginning.
F# uses the colon (:) notation to indicate types of things. Let's say you define a value of type int:
let myNumber = 5
F# Interactive will understand that myNumber is an integer, and will tell you this by:
myNumber : int
which is read as
myNumber is of type int
F# functional types
So far so good. Let's introduce something else, functional types. A functional type is simply the type of a function. F# uses -> to denote a functional type. This arrow symbolizes that what is written on its left-hand side is transformed into what is written into its right-hand side.
Let's consider a simple function, that takes one argument and transforms it into one output. An example of such a function would be:
isEven : int -> bool
This introduces the name of the function (on the left of the :), and its type. This line can be read in English as:
isEven is of type function that transforms an int into a bool.
Note that to correctly interpret what is being said, you should make a short pause just after the part "is of type", and then read the rest of the sentence at once, without pausing.
In F# functions are values
In F#, functions are (almost) no more special than ordinary types. They are things that you can pass around to functions, return from functions, just like bools, ints or strings.
So if you have:
myNumber : int
isEven : int -> bool
You should consider int and int -> bool as two entities of the same kind: types. Here, myNumber is a value of type int, and isEven is a value of type int -> bool (this is what I'm trying to symbolize when I talk about the short pause above).
Function application
Values of types that contain -> happens to be also called functions, and have special powers: you can apply a function to a value. So, for example,
isEven myNumber
means that you are applying the function called isEven to the value myNumber. As you can expect by inspecting the type of isEven, it will return a boolean value. If you have correctly implemented isEven, it would obviously return false.
A function that returns a value of a functional type
Let's define a generic function to determine is an integer is multiple of some other integer. We can imagine that our function's type will be (the parenthesis are here to help you understand, they might or might not be present, they have a special meaning):
isMultipleOf : int -> (int -> bool)
As you can guess, this is read as:
isMultipleOf is of type (PAUSE) function that transforms an int into (PAUSE) function that transforms an int into a bool.
(here the (PAUSE) denote the pauses when reading out loud).
We will define this function later. Before that, let's see how we can use it:
let isEven = isMultipleOf 2
F# interactive would answer:
isEven : int -> bool
which is read as
isEven is of type int -> bool
Here, isEven has type int -> bool, since we have just given the value 2 (int) to isMultipleOf, which, as we have already seen, transforms an int into an int -> bool.
We can view this function isMultipleOf as a sort of function creator.
Definition of isMultipleOf
So now let's define this mystical function-creating function.
let isMultipleOf n x =
(x % n) = 0
Easy, huh?
If you type this into F# Interactive, it will answer:
isMultipleOf : int -> int -> bool
Where are the parenthesis?
Note that there are no parenthesis. This is not particularly important for you now. Just remember that the arrows are right associative. That is, if you have
a -> b -> c
you should interpret it as
a -> (b -> c)
The right in right associative means that you should interpret as if there were parenthesis around the rightmost operator. So:
a -> b -> c -> d
should be interpreted as
a -> (b -> (c -> d))
Usages of isMultipleOf
So, as you have seen, we can use isMultipleOf to create new functions:
let isEven = isMultipleOf 2
let isOdd = not << isEven
let isMultipleOfThree = isMultipleOf 3
let endsWithZero = isMultipleOf 10
F# Interactive would respond:
isEven : int -> bool
isOdd : int -> bool
isMultipleOfThree : int -> bool
endsWithZero : int -> bool
But you can use it differently. If you don't want to (or need to) create a new function, you can use it as follows:
isMultipleOf 10 150
This would return true, as 150 is multiple of 10. This is exactly the same as create the function endsWithZero and then applying it to the value 150.
Actually, function application is left associative, which means that the line above should be interpreted as:
(isMultipleOf 10) 150
That is, you put the parenthesis around the leftmost function application.
Now, if you can understand all this, your example (which is the canonical CreateAdder) should be trivial!
Sometime ago someone asked this question which deals with exactly the same concept, but in Javascript. In my answer I give two canonical examples (CreateAdder, CreateMultiplier) inf Javascript, that are somewhat more explicit about returning functions.
I hope this helps.
The canonical example of this is probably an "adder creator" - a function which, given a number (e.g. 3) returns another function which takes an integer and adds the first number to it.
So, for example, in pseudo-code
x = CreateAdder(3)
x(5) // returns 8
x(10) // returns 13
CreateAdder(20)(30) // returns 50
I'm not quite comfortable enough in F# to try to write it without checking it, but the C# would be something like:
public static Func<int, int> CreateAdder(int amountToAdd)
{
return x => x + amountToAdd;
}
Does that help?
EDIT: As Bruno noted, the example you've given in your question is exactly the example I've given C# code for, so the above pseudocode would become:
let x = f1 3
x 5 // Result: 8
x 10 // Result: 13
f1 20 30 // Result: 50
It's a function that takes an integer and returns a function that takes an integer and returns an integer.
This is functionally equivalent to a function that takes two integers and returns an integer. This way of treating functions that take multiple parameters is common in functional languages and makes it easy to partially apply a function on a value.
For example, assume there's an add function that takes two integers and adds them together:
let add x y = x + y
You have a list and you want to add 10 to each item. You'd partially apply add function to the value 10. It would bind one of the parameters to 10 and leaves the other argument unbound.
let list = [1;2;3;4]
let listPlusTen = List.map (add 10)
This trick makes composing functions very easy and makes them very reusable. As you can see, you don't need to write another function that adds 10 to the list items to pass it to map. You have just reused the add function.
You usually interpret this as a function that takes two integers and returns an integer.
You should read about currying.
a function taking an integer, which returns a function which takes an integer and returns an integer
The last part of that:
a function which takes an integer and returns an integer
should be rather simple, C# example:
public int Test(int takesAnInteger) { return 0; }
So we're left with
a function taking an integer, which returns (a function like the one above)
C# again:
public int Test(int takesAnInteger) { return 0; }
public int Test2(int takesAnInteger) { return 1; }
public Func<int,int> Test(int takesAnInteger) {
if(takesAnInteger == 0) {
return Test;
} else {
return Test2;
}
}
You may want to read
F# function types: fun with tuples and currying
In F# (and many other functional languages), there's a concept called curried functions. This is what you're seeing. Essentially, every function takes one argument and returns one value.
This seems a bit confusing at first, because you can write let add x y = x + y and it appears to add two arguments. But actually, the original add function only takes the argument x. When you apply it, it returns a function that takes one argument (y) and has the x value already filled in. When you then apply that function, it returns the desired integer.
This is shown in the type signature. Think of the arrow in a type signature as meaning "takes the thing on my left side and returns the thing on my right side". In the type int -> int -> int, this means that it takes an argument of type int — an integer — and returns a function of type int -> int — a function that takes an integer and returns an integer. You'll notice that this precisely matches the description of how curried functions work above.
Example:
let f b a = pown a b //f a b = a^b
is a function that takes an int (the exponent) and returns a function that raises its argument to that exponent, like
let sqr = f 2
or
let tothepowerofthree = f 3
so
sqr 5 = 25
tothepowerofthree 3 = 27
The concept is called Higher Order Function and quite common to functional programming.
Functions themselves are just another type of data. Hence you can write functions that return other functions. Of course you can still have a function that takes an int as parameter and returns something else. Combine the two and consider the following example (in python):
def mult_by(a):
def _mult_by(x):
return x*a
return mult_by
mult_by_3 = mult_by(3)
print mylt_by_3(3)
9
(sorry for using python, but i don't know f#)
There are already lots of answers here, but I'd like to offer another take. Sometimes explaining the same thing in lots of different ways helps you to 'grok' it.
I like to think of functions as "you give me something, and I'll give you something else back"
So a Func<int, string> says "you give me an int, and I'll give you a string".
I also find it easier to think in terms of 'later' : "When you give me an int, I'll give you a string". This is especially important when you see things like myfunc = x => y => x + y ("When you give curried an x, you get back something which when you give it a y will return x + y").
(By the way, I'm assuming you're familiar with C# here)
So we could express your int -> int -> int example as Func<int, Func<int, int>>.
Another way that I look at int -> int -> int is that you peel away each element from the left by providing an argument of the appropriate type. And when you have no more ->'s, you're out of 'laters' and you get a value.
(Just for fun), you can transform a function which takes all it's arguments in one go into one which takes them 'progressively' (the official term for applying them progressively is 'partial application'), this is called 'currying':
static void Main()
{
//define a simple add function
Func<int, int, int> add = (a, b) => a + b;
//curry so we can apply one parameter at a time
var curried = Curry(add);
//'build' an incrementer out of our add function
var inc = curried(1); // (var inc = Curry(add)(1) works here too)
Console.WriteLine(inc(5)); // returns 6
Console.ReadKey();
}
static Func<T, Func<T, T>> Curry<T>(Func<T, T, T> f)
{
return a => b => f(a, b);
}
Here is my 2 c. By default F# functions enable partial application or currying. This means when you define this:
let adder a b = a + b;;
You are defining a function that takes and integer and returns a function that takes an integer and returns an integer or int -> int -> int. Currying then allows you partiallly apply a function to create another function:
let twoadder = adder 2;;
//val it: int -> int
The above code predifined a to 2, so that whenever you call twoadder 3 it will simply add two to the argument.
The syntax where the function parameters are separated by space is equivalent to this lambda syntax:
let adder = fun a -> fun b -> a + b;;
Which is a more readable way to figure out that the two functions are actually chained.