I noticed that I get the same effect if I define this trivial function:
fun double ( i: Int ) = i*2
and if I define a variable and assign a lambda (with an identical body) to it:
var double = { i : Int -> i*2 }
I get the same result if I call double(a) with either declaration.
This leaves me confused. When is it needed, recommended, advantageous to define a variable as a lambda rather than define a function to it?
When is it needed, recommended, advantageous to define a variable as a lambda rather than define a function to it?
Whenever you have the choice of either, you should use a fun declaration. Even with a fun you can still get a first-class callable object from it by using a function reference.
On the JVM, a fun is significantly more lightweight, both in terms of RAM and invocation overhead. It compiles into a Java method, whereas a val compiles into an instance field + getter + a synthetic class that implements a functional interface + a singleton instance of that class that you must fetch, dereference, and invoke a method on it.
You should consider a function-typed val or var only when something is forcing you to do it. One example is that you can dynamically replace a var and effectively change the definition of the function. You may also receive function objects from the outside, or you may need to comply with an API that needs them.
In any case, if you ever use a function-typed property of a class, you'll know why you're doing it.
First, if I understand you right, your question is "Why are functions first-class citizens in Kotlin -- And when to use them as such?", right?
Kotlin functions are first-class, which means that they can be stored in variables and data structures, passed as arguments to and returned from other higher-order functions. You can operate with functions in any way that is possible for other non-function values. (see here)
As stated in the docs, one use case are higher-order functions. As a first step, I will leave the wikipedia link here: https://en.wikipedia.org/wiki/Higher-order_function
Basically, a higher-order function is a function that takes functions as parameters, or returns a function.
This means that a higher-order function has at least one parameter of a function type or returns a value of a function type.
Following a short example of a higher-order function that receives a parameter of function type (Int) -> Boolean:
fun foo(pred: (Int) -> Boolean) : String = if(pred(x)) "SUCCESS" else "FAIL"
This higher-order function can now be called with any (Int) -> Boolean function.
The docs also state ... [can be used] in any way that is possible for other non-function values.
This means that you can, for example, assign different functions to a variable, depending on your current context.
For example:
// This example is verbose on purpose ;)
var checker: (Int) -> Boolean
if (POSITIVE_CHECK) {
checker = { x -> x > 0 } // Either store this function ...
} else {
checker = { x -> x < 0 } // ... or this one ...
}
if (checker(someNumber)) { // ... and use whatever function is now stored in variable "checker" here
print("Check was fine")
}
(Code untested)
You can define variable and assign it lambda when you want change behaviour for some reason. For example, you have different formula for several cases.
val formula: (Int) -> Int = when(value) {
CONDITION1 -> { it*2 }
CONDITION2 -> { it*3 }
else -> { it }
}
val x: Int = TODO()
val result = formula(x)
If you simply need helper function, you should define it as fun.
If you pass a lambda as a parameter of a function it will be stored in a variable. The calling application might need to save that (e.g. event listener for later use). Therefore you need to be able to store it as a variable as well. As said in the answer however, you should do this only when needed!
For me, I would write the Lambda variable as followed:
var double: (Int) -> Int = { i -> //no need to specify parameter name in () but in {}
i*2
}
So that you can easily know that its type is (i: Int) -> Int, read as takes an integer and returns an integer.
Then you can pass it to somewhere say a function like:
fun doSomething(double: (Int) -> Int) {
double(i)
}
Related
I can define a generic function:
package hello
func IsZero[T int64|float64](value T) bool {
return value == 0
}
Then if I try to alias that function in another package, it fails:
package world
import "hello"
var IsZero = hello.IsZero
The above doesn't compile with:
cannot use generic function hello.IsZero without instantiation
Instead this works:
var IsZero = hello.IsZero[int64]
Is it possible to do this, using some other syntax?
That's not an alias. And you already have your answer, actually. But if you want a formal reference, from the language specs, Instantiations:
A generic function that is is not called requires a type argument list for instantiation
So when you attempt to initialize a variable of function type, the function hello.IsZero is not called, and therefore requires instantiation with specific type parameters:
// not called, instantiated with int64
var IsZero = hello.IsZero[int64]
At this point the variable (let's give it a different name for clarity) zeroFunc has a concrete function type:
var zeroFunc = IsZero[int64]
fmt.Printf("type: %T\n", zeroFunc)
Prints:
type: func(int64) bool
This might or might not be what you want, as this effectively monomorphises the function.
If you just want to have a local symbol, with the same implementation (or a tweaked version thereof), declaring a "wrapper" function works. Just remember that the type parameters of your wrapper can only be as strict or stricter than the wrapped one's
E.g. Given
IsZero[T int64 | float64](v T)
your wrapper can not be
WrapIsZeroPermissive[T int64 | float64 | complex128](v T) bool {
return IsZero(v) // does not compile, T's type set is a superset
}
but can be
WrapIsZeroStricter[T int64](v T) bool {
return IsZero(v) // ok, T's type set is a subset
}
If the function is small, like in the question, it's probably easier to just vendor it:
package vendor
func thisIsJustCopy[T int64|float64](value T) bool {
return value == 0
}
but if the function is big, you can do it like this:
package world
import "hello"
func IsZero[T int64|float64](value T) bool {
return hello.IsZero(value)
}
I try to alias that function in another package
Aliases work for types only. Your code just tries to declare a variable.
Is it possible to do this, using some other syntax?
No.
I think an extension lambda requires that you pass in the correct argument, but i seems not to do so in the following example.
open class Base {
open fun f() = 1
}
class Derived : Base() {
override fun f() = 99
}
fun Base.g(): Int { return f()}
fun Base.h(xl: Base.() -> Int): Int { return xl()}
fun main() {
val b: Base = Derived() // Upcast
println(b.g())
println(b.h { f()}) // [1]
}
I understand that Base.h takes a function that takes a Base object as its parameter. But line [1] shows that it accepts f(), which is a function that takes no parameter. I was thinking hard about this and I prefixed it with this.f() and it still worked. Not convinced, I modified the code as follows:
open class Base {
open fun f() = 1
}
class Derived : Base() {
override fun f() = 99
}
fun Base.g(): Int { return f()}
fun Base.h(xl: (Base) -> Int): Int { return xl(Base())}
fun test(i:Int) = 1
fun main() {
val b: Base = Derived() // Upcast
println(b.g())
println(b.h { test(1) })
}
This code works. I've run it to verify. And as you can see, b.h() accepts test(), which takes an Int. And this is contrary to the fact that Base.h() takes a Base.
Could you explain this? Thank you for reading.
Note the curly brackets around the functions that are passed in! They change everything.
In the second code, b.h { test(1) } is not passing the function test to b.h. The syntax to pass test to b.h would be b.h(::test), and that does produce an error as you would expect.
b.h { test(1) } passes a function (a lambda expression) that takes a Base as parameter, ignores that parameter, calls test(1) and returns the result. You are basically passing a function that looks like this to b.h:
fun foo(p: Base) = test(1)
You might be wondering how Kotlin knows about Base when you did not write the word Base in the call at all. Well, it can just look at the declaration of b.h, and see that { test(1) } must take a parameter of Base.
The first code snippet is a bit different, because b.h accepts a Base.() -> Int in that case. Base.() -> Int represents a function whose receiver type is Base, that is, a function func that can be called like someBaseObject.func(). Compare this to a function func that takes a Base object as parameter, which can be called like func(someBaseObject).
Again, { f() } is not passing the function f. It is a lambda expression that does nothing but calls f. In this case though, f itself can be passed to b.h (b.h(Base::f)), because it is a function with a receiver type of Base! You can do someBaseObject.f(), can't you? Passing the lambda is similar to passing an extension function that is declared like this (you're just "wrapping" f in another function):
fun Base.foo() = f()
And since the receiver of the function is Base, you are able to access other functions that has Base as the receiver (such as f) in the lambda. You can also specify the receiver (which is this) explicitly.
Lets assume you have a class:
class Person{
var age :Int?
var name :String?
}
and then you have collection of this class, let's call it people :List.
You can call sort as:
people.sortBy{person -> person.age}
or
people.sortBy{person -> person.name}
Im wondering if its possible to write function definition that would sort by a given field? such as:
fun sortbyField(field:???){
peple.sortBy{field}
}
I have no idea if its possible, if so, how to define "field" parameter.
Thanks!
You can make the function parameter accept a function type with a receiver, like Person.() -> T, and then, inside the bodies of lambdas passed to the function, it will be possible to access a property of the implicit receiver:
fun <T : Comparable<T>> sortUsing(fn: Person.() -> T) {
people.sortBy { it.fn() }
}
Usages:
sortUsing { name }
sortUsing { age }
Alternatively, you can pass a callable reference to the property as a functional argument, instead of a lambda:
people.sortBy(Person::name)
people.sortBy(Person::age)
This works for both functional parameters accepting a single argument, (Person) -> T, and for functional parameters with receiver, Person.() -> T.
You can simply use the field as a parameter if you do not call the function as a lambda expression:
people.sortedBy(Person::name)
Maybe the following is something you were looking for:
fun List<Person>.sortByHeader(header: String) = sortedWith(
when (header) {
"header_name" -> compareBy(nullsFirst(), Person::name)
"header_age" -> compareBy(nullsLast(), Person::age)
"header_time" -> compareBy(Person::time)
else -> compareBy(Person::id)
}
)
I used the following data class instead:
data class Person(val id: Int, val name: String?, val age : Int?, val time: LocalDateTime)
This way you can then call it using your header name, e.g.:
pepple.sortByHeader("header_id").run(::println)
pepple.sortByHeader("header_name").run(::println)
pepple.sortByHeader("header_time").run(::println)
pepple.sortByHeader("header_age").run(::println)
I just added some nullsFirst/nullsLast in case you want to have something like that in place as well.
If that is not what you were after, then using a function with receiver as shown by hotkeys answer might be more appropriate for you. The next alternative is using reflection, but I will omit this one, as that should be used rather as a last resort.
Yet another way to write it
people.sortBy { it.name }
In kotlin, when a lambda expect only one argument, you can simplify the syntax: people.sort { it -> it.name } becomes people.sort { it.name }. You have to use the name it though
In the following code, I'm trying to create a "function pointer" and an array of functions by regarding function names as usual variables:
proc myfunc1() { return 100; }
proc myfunc2() { return 200; }
// a function variable?
var myfunc = myfunc1;
writeln( myfunc() );
myfunc = myfunc2;
writeln( myfunc() );
// an array of functions?
var myfuncs: [1..2] myfunc1.type;
writeln( myfuncs.type: string );
myfuncs[ 1 ] = myfunc1;
myfuncs[ 2 ] = myfunc2;
for fun in myfuncs do
writeln( fun() );
which seems to be working as expected (with Chapel v1.16)
100
200
[domain(1,int(64),false)] chpl__fcf_type_void_int64_t
100
200
So I'm wondering whether the above usage of function variables is legitimate? For creating an array of functions, is it usual to define a concrete function with desired signature first and then refer to its type (with .type) as in the above example?
Also, is it no problem to treat such variables as "usual" variables, e.g., pass them to other functions as arguments or include them as a field of class/record? (Please ignore these latter questions if they are too broad...) I would appreciate any advice if there are potential pitfalls (if any).
This code is using first class function support, which is prototype/draft in the Chapel language design. You can read more about the prototype support in the First-class Functions in Chapel technote.
While many uses of first-class functions work in 1.16 and later versions, you can expect that the language design in this area will be revisited. In particular there isn't currently a reasonable answer to the question of whether or not variables can be captured (and right now attempting to do so probably results in a confusing error). I don't know in which future release this will change, though.
Regarding the myfunc1.type part, the section in the technote I referred to called "Specifying the type of a first-class function" presents an alternative strategy. However I don't see any problem with using myfunc1.type in this case.
Lastly, note that the lambda support in the current compiler actually operates by creating a class with a this method. So you can do the same - create a "function object" (to borrow a C++ term) - that has the same effect. A "function object" could be a record or a class. If it's a class, you might use inheritance to be able to create an array of objects that can respond to the same method depending on their dynamic type. This strategy might allow you to work around current issues with first class functions. Even if first-class-function support is completed, the "function object" approach allow you to be more explicit about captured variables. In particular, you might store them as fields in the class and set them in the class initializer. Here is an example creating and using an array of different types of function objects:
class BaseHandler {
// consider these as "pure virtual" functions
proc name():string { halt("base name called"); }
proc this(arg:int) { halt("base greet called"); }
}
class HelloHandler : BaseHandler {
proc name():string { return "hello"; }
proc this(arg:int) { writeln("Hello ", arg); }
}
class CiaoHandler : BaseHandler {
proc name():string { return "ciao"; }
proc this(arg:int) { writeln("Ciao ", arg); }
}
proc test() {
// create an array of handlers
var handlers:[1..0] BaseHandler;
handlers.push_back(new HelloHandler());
handlers.push_back(new CiaoHandler());
for h in handlers {
h(1); // calls 'this' method in instance
}
}
test();
Yes, in your example you are using Chapel's initial support for first-class functions. To your second question, you could alternatively use a function type helper for the declaration of the function array:
var myfuncs: [1..2] func(int);
These first-class function objects can be passed as arguments into functions – this is how Futures.async() works – or stored as fields in a record (Try It Online! example). Chapel's first-class function capabilities also include lambda functions.
To be clear, the "initial" aspect of this support comes with the caveat (from the documentation):
This mechanism should be considered a stopgap technology until we have developed and implemented a more robust story, which is why it's being described in this README rather than the language specification.
I am reading about boost::function and I am a bit confused about its use and its relation to other C++ constructs or terms I have found in the documentation, e.g. here.
In the context of C++ (C++11), what is the difference between an instance of boost::function, a function object, a functor, and a lambda expression? When should one use which construct? For example, when should I wrap a function object in a boost::function instead of using the object directly?
Are all the above C++ constructs different ways to implement what in functional languages is called a closure (a function, possibly containing captured variables, that can be passed around as a value and invoked by other functions)?
A function object and a functor are the same thing; an object that implements the function call operator operator(). A lambda expression produces a function object. Objects with the type of some specialization of boost::function/std::function are also function objects.
Lambda are special in that lambda expressions have an anonymous and unique type, and are a convenient way to create a functor inline.
boost::function/std::function is special in that it turns any callable entity into a functor with a type that depends only on the signature of the callable entity. For example, lambda expressions each have a unique type, so it's difficult to pass them around non-generic code. If you create an std::function from a lambda then you can easily pass around the wrapped lambda.
Both boost::function and the standard version std::function are wrappers provided by the library. They're potentially expensive and pretty heavy, and you should only use them if you actually need a collection of heterogeneous, callable entities. As long as you only need one callable entity at a time, you are much better off using auto or templates.
Here's an example:
std::vector<std::function<int(int, int)>> v;
v.push_back(some_free_function); // free function
v.push_back(&Foo::mem_fun, &x, _1, _2); // member function bound to an object
v.push_back([&](int a, int b) -> int { return a + m[b]; }); // closure
int res = 0;
for (auto & f : v) { res += f(1, 2); }
Here's a counter-example:
template <typename F>
int apply(F && f)
{
return std::forward<F>(f)(1, 2);
}
In this case, it would have been entirely gratuitous to declare apply like this:
int apply(std::function<int(int,int)>) // wasteful
The conversion is unnecessary, and the templated version can match the actual (often unknowable) type, for example of the bind expression or the lambda expression.
Function Objects and Functors are often described in terms of a
concept. That means they describe a set of requirements of a type. A
lot of things in respect to Functors changed in C++11 and the new
concept is called Callable. An object o of callable type is an
object where (essentially) the expression o(ARGS) is true. Examples
for Callable are
int f() {return 23;}
struct FO {
int operator()() const {return 23;}
};
Often some requirements on the return type of the Callable are added
too. You use a Callable like this:
template<typename Callable>
int call(Callable c) {
return c();
}
call(&f);
call(FO());
Constructs like above require you to know the exact type at
compile-time. This is not always possible and this is where
std::function comes in.
std::function is such a Callable, but it allows you to erase the
actual type you are calling (e.g. your function accepting a callable
is not a template anymore). Still calling a function requires you to
know its arguments and return type, thus those have to be specified as
template arguments to std::function.
You would use it like this:
int call(std::function<int()> c) {
return c();
}
call(&f);
call(FO());
You need to remember that using std::function can have an impact on
performance and you should only use it, when you are sure you need
it. In almost all other cases a template solves your problem.