I am following this tutorial and I see this code:
println("\nStep 1: How to define a higher order function which takes another function as parameter")
def totalCostWithDiscountFunctionParameter(donutType: String)(quantity: Int)(f: Double => Double): Double = {
println(s"Calculating total cost for $quantity $donutType")
val totalCost = 2.50 * quantity
f(totalCost)
}
println("\nStep 2: How to define and pass a def function to a higher order function")
def applyDiscount(totalCost: Double): Double = {
val discount = 2 // assume you fetch discount from database
totalCost - discount
}
println(s"Total cost of 5 Glazed Donuts with discount def function = ${totalCostWithDiscountFunctionParameter("Glazed Donut")(5)(applyDiscount(_))}")
println("\nStep 3: How to define and pass a val function to a higher order function")
val applyDiscountValueFunction = (totalCost: Double) => {
val discount = 2 // assume you fetch discount from database
totalCost - discount
}
println(s"Total cost of 5 Glazed Donuts with discount val function = ${totalCostWithDiscountFunctionParameter("Glazed Donut")(5)(applyDiscountValueFunction)}")
The author says:
The function has a by-name parameter which is expected to be a function that has a parameter of type Double and will also return a type of Double.
Is that true? What would be a by-value parameter here? Is the function lazily evaluated which makes it a by-name parameter? Is that true?
Why does the author use the wildcard when passing in applyDiscount but not when passing in applydiscountValueFunction? Both work without the wildcard operator.
applyDiscount(_) transforms the method into a function, using placeholder syntax for anonymous functions.
This process can be automated by the compiler (using a technique called eta-expansion) whenever a function type is expected and a method is passed instead, which is exactly the case in the example.
(for a more thorough discussion, refer to this answer: Underscore usage when passing a function in Scala)
So you're right in saying that this
totalCostWithDiscountFunctionParameter("Glazed Donut")(5)(applyDiscount)
would work anyway, because the compiler will transform the applyDiscount to a function automatically.
As per the by-name parameter, what the author is calling by-name parameter (the parameter f) is really just a parameter of type Double => Double, so it seems he's using incorrect terminology.
A by-name parameter in Scala is expressed using => like this:
def someMethod(someParam: => Double): Double = // ...
In this example someParam is a by name parameter, meaning it won't be evaluated at call-site.
Related
I have a method that with the implicit parameter. i get a error when i convert it to function in 2 case :
1:
def action(implicit i:Int) = i + " in action"
val f = action _
then i get a StackOverflowError.
2：
def action(implicit i:Int) = i + " in action"
val f = action(_)
then i get a error: missing parameter type
I must write like this :
val f = (i:Int) => action(i)
that's ok. And if the parameter of 'action' is not the implicit , all case are right. So how to explain , and what i miss ?
If you specify a parameter to a function to be implicit, you are inviting the compiler to supply the value of that parameter for you. So how does the compiler find those values? It looks for values of the same type (Int in your case) that have been declared as implicit values in a variety of scopes.
(For simplicity, I'll just use a local scope in this example, but you might want to read up on this topic. Programming in Scala, 3rd Ed is a good first step.)
Note that the names of the implicit values are ignored and have no bearing on proceedings, the compiler only looks at the types of implicit values. If multiple implicit values with the required type are found in the same scope, then the compiler will complain about ambiguous implicit values.
For example, the following provides a function with an implicit parameter and a default value for that parameter within the current scope:
def greetPerson(name: String)(implicit greeting: String) = s"$greeting $name!"
implicit val defaultGreeting = "Hello" // Implicit value to be used for greeting argument.
val y = greetPerson("Bob") // Equivalent to greetPerson("Bob")(defaultGreeting).
val z = greetPerson("Fred")("Hi")
Note that y is just a String value of "Hello Bob!", and z is a string with the value "Hi Fred!"; neither of them are functions.
Also note that greetPerson is a curried function. This is because implicit parameters cannot be mixed with regular, non-implicit parameters in the same parameter list.
In general, it's bad practice to use common types (Int, Boolean, String, etc.) as values for implicit parameters. In a big program, there might be a lot of different implicit values in your scope, and you might pick up an unexpected value. For that reason, it's standard practice to wrap such values in a case class instead.
If you're trying to create a value that supplies some of the arguments of another function (that is, a partially applied function), then that would look something like this:
def greetPerson(greeting: String, name: String) = s"$greeting $name!"
val sayHello = greetPerson("Hello", _: String)
val y = sayHello("Bob") // "Hello Bob!"
val sayHi = greetPerson("Hi", _: String)
val z = sayHi("Fred") // "Hi Fred!"
In both cases, we're creating partially applied functions (sayHi and sayHello) that call greetPerson with the greeting parameter specified, but which allow us to specify the name parameter. Both sayHello and sayHi are still only values, but their values are partially applied functions rather than constants.
Depending upon your circumstances, I think the latter case may suit you better...
I would also read up on how the underscore character (_) is used in Scala. In a partially applied function declaration, it corresponds to the arguments that will be provided later. But it has a lot of other uses too. I think there's no alternative to reading up on Scala and learning how and when to use them.
I've written a simple code in Scala with implicit conversion of Function1 to some case class.
object MyApp extends App{
case class FunctionContainer(val function:AnyRef)
implicit def cast(function1: Int => String):FunctionContainer = new FunctionContainer(function1)
def someFunction(i:Int):String = "someString"
def abc(f : FunctionContainer):String = "abc"
println(abc(someFunction))
}
But it doesn't work. Compiler doesn't want to pass someFunction as an argument to abc. I can guess its reasons but don't know exactly why it doesn't work.
When you use a method name as you have, the compiler has to pick how to convert the method type to a value. If the expected type is a function, then it eta-expands; otherwise it supplies empty parens to invoke the method. That is described here in the spec.
But it wasn't always that way. Ten years ago, you would have got your function value just by using the method name.
The new online spec omits the "Change Log" appendix, so for the record, here is the moment when someone got frustrated with parens and introduced the current rules. (See Scala Reference 2.9, page 181.)
This has not eliminated all irksome anomalies.
Conversions
The rules for implicit conversions of methods to functions (§6.26) have been tightened. Previously, a parameterized method used as a value was always implicitly converted to a function. This could lead to unexpected results when method arguments were forgotten. Consider for instance the statement below:
show(x.toString)
where show is defined as follows:
def show(x: String) = Console.println(x)
Most likely, the programmer forgot to supply an empty argument list () to toString. The previous Scala version would treat this code as a partially applied method, and expand it to:
show(() => x.toString())
As a result, the address of a closure would be printed instead of the value of s. Scala version 2.0 will apply a conversion from partially applied method to function value only if the expected type of the expression is indeed a function type. For instance, the conversion would not be applied in the code above because the expected type of show’s parameter is String, not a function type. The new convention disallows some previously legal code. Example:
def sum(f: int => double)(a: int, b: int): double =
if (a > b) 0 else f(a) + sum(f)(a + 1, b)
val sumInts = sum(x => x) // error: missing arguments
The partial application of sum in the last line of the code above will not be converted to a function type. Instead, the compiler will produce an error message which states that arguments for method sum are missing. The problem can be fixed by providing an expected type for the partial application, for instance by annotating the definition of sumInts with its type:
val sumInts: (int, int) => double = sum(x => x) // OK
On the other hand, Scala version 2.0 now automatically applies methods with empty parameter lists to () argument lists when necessary. For instance, the show expression above will now be expanded to
show(x.toString())
Your someFunction appears as a method here.
You could try either
object MyApp extends App{
case class FunctionContainer(val function:AnyRef)
implicit def cast(function1: Int => String):FunctionContainer = new FunctionContainer(function1)
val someFunction = (i:Int) => "someString"
def abc(f : FunctionContainer):String = "abc"
println(abc(someFunction))
}
or
object MyApp extends App{
case class FunctionContainer(val function:AnyRef)
implicit def cast(function1: Int => String):FunctionContainer = new FunctionContainer(function1)
def someFunction(i:Int): String = "someString"
def abc(f : FunctionContainer):String = "abc"
println(abc(someFunction(_: Int)))
}
By the way: implicitly casting such common functions to something else can quickly lead to problems. Are you absolutely sure that you need this? Wouldn't it be easier to overload abc?
You should use eta-expansion
println(abc(someFunction _))
I am really new to Scala and trying to study it.
I don't know how to access or using the parameter of higher order function. For example:
def higherOrderFunc(f: Int => Boolean): String = {
/* Logic to print parameter is here */
"Hello"
}
val func = higherOrderFunc(x => x > 1)
How can I print the value of x before I return value "Hello"
You can't. The argument does not exist in this context; it'd need to be passed into the higher-order function along with the anonymous function.
I have a loan pattern that applies a function n times where 'i' is the incrementing variable. "Occasionally", I want the function passed in to have access to 'i'....but I don't want to require all functions passed in to require defining a param to accept 'i'. Example below...
def withLoaner = (n:Int) => (op:(Int) => String) => {
val result = for(i <- 1 to n) yield op(i)
result.mkString("\n")
}
def bob = (x:Int) => "bob" // don't need access to i. is there a way use () => "bob" instead?
def nums = (x:Int) => x.toString // needs access to i, define i as an input param
println(withLoaner(3)(bob))
println(withLoaner(3)(nums))
def withLoaner(n: Int) = new {
def apply(op: Int => String) : String = (1 to n).map(op).mkString("\n")
def apply(op: () => String) : String = apply{i: Int => op()}
}
(not sure how it is related to the loan pattern)
Edit Little explanation as requested in comment.
Not sure what you know and don't know of scala and what you don't undestand in that code. so sorry if what I just belabor the obvious.
First, a scala program consist of traits/classes (also singleton object) and methods. Everything that is done is done by methods (leaving constructor aside). Functions (as opposed to methods) are instances of (subtypes of) the various FunctionN traits (N the number of arguments). Each of them has as apply method that is the actual implemention.
If you write
val inc = {i: Int => i + 1}
it is desugared to
val inc = new Function1[Int, Int] {def apply(i: Int) = i + 1}
(defines an anonymous class extending Function1, with given apply method and creating an instance)
So writing a function has rather more weight than a simple method. Also you cannot have overloading (several methods with the same name, differing by the signature, just what I did above), nor use named arguments, or default value for arguments.
On the other hand, functions are first classes values (they can be passed as arguments, returned as result) while methods are not. They are automatically converted to functions when needed, however there may be some edges cases when doing that. If a method is intended solely to be used as a function value, rather than called as a method, it might be better to write a function.
A function f, with its apply method, is called with f(x) rather than f.apply(x) (which works too), because scala desugars function call notation on a value (value followed by parentheses and 0 or more args) to a call to method apply. f(x) is syntactic sugar for f.apply(x). This works whatever the type of f, it does not need to be one of the FunctionN.
What is done in withLoaner is returning an object (of an anonymous type, but one could have defined a class separately and returned an instance of it). The object has two apply methods, one accepting an Int => String, the other one an () => String. When you do withLoaner(n)(f) it means withLoaner(n).apply(f). The appropriate apply method is selected, if f has the proper type for one of them, otherwise, compile error.
Just in case you wonder withLoaner(n) does not mean withLoaner.apply(n) (or it would never stop, that could just as well mean withLoaner.apply.apply(n)), as withLoaner is a method, not a value.
def func(arg: String => Int): Unit = {
// body of function
}
I mean this fragment:
String => Int
Short answer
Its a function that receives a String and returns a Int
Long answer
In Scala, functions are first class citizens. That means you can store them in variables or (like in this case) pass them around as arguments.
This is how a function literal looks like
() => Unit
This is a function that receives no arguments and returns Unit (java's equivalent to void).
This would be a function that receives a String as a parameter and returns an Int:
(String) => Int
Also, scala let's you drop the parenthesis as a form of syntactic sugar, like in your example. The preceding arg: is just the name of the argument.
Inside func you would call the function received (arg) like this:
val result = arg("Some String") // this returns a Int
As mentioned in Advantages of Scala’s Type System, it is a Functional type.
The article Scala for Java Refugees Part 6: Getting Over Java describes this syntax in its section "Higher-Order Functions".
def itrate(array:Array[String], fun:(String)=>Unit) = {
for (i <- 0 to (array.length - 1)) { // anti-idiom array iteration
fun(array(i))
}
}
val a = Array("Daniel", "Chris", "Joseph", "Renee")
iterate(a, (s:String) => println(s))
See? The syntax is so natural you almost miss it.
Starting at the top, we look at the type of the fun parameter and we see the (type1, …)=>returnType syntax which indicates a functional type.
In this case, fun will be a functional which takes a single parameter of type String and returns Unit (effectively void, so anything at all).
Two lines down in the function, we see the syntax for actually invoking the functional. fun is treated just as if it were a method available within the scope, the call syntax is identical.
Veterans of the C/C++ dark-ages will recognize this syntax as being reminiscent of how function pointers were handled back-in-the-day.
The difference is, no memory leaks to worry about, and no over-verbosity introduced by too many star symbols.
In your case: def func(arg: String => Int): Unit, arg would be a function taking a String and returning an Int.
You might also see it written (perhaps by a decompiler) as
def func(arg: Function1[String, Int]): Unit = {
// body of function
}
They are precisely equivalent.