I realize the following function calls are all same, but I do not understand why.
val list = List(List(1), List(2, 3), List(4, 5, 6))
list.map(_.length) // res0 = List(1,2,3) result of 1st call
list map(_.length) // res1 = List(1,2,3) result of 2nd call
list map (_.length) // res2 = List(1,2,3) result of 3rd call
I can understand 1st call, which is just a regular function call because map is a member function of class List
But I can not understand 2nd and 3rd call. For example, in the 3rd call, how can Scala compiler know "(_.length)" is parameter of "map"? How can compiler know "map" is a member function of "list"?
The only difference between variant 2 and 3 is the blank in front of the parenthesis? This can only be a delimiter - list a and lista is of course different, but a opening parens is a new token, and you can put a blank or two or three in front - or none. I don't see how you can expect a difference here.
In Java, there is no difference between
System.out.println ("foo");
// and
System.out.println("foo");
too.
This is the operator notation. The reason it works is the same reason why 2 + 2 works.
The space is used to distinguish between words -- listmap(_.length) would make the compiler look for listmap. But if you write list++list, it will work too, as will list ++ list.
So, one you are using operator notation, the space is necessary to separate words, but otherwise may be present or not.
Related
I'd like the example computation expression and values below to return 6. For some the numbers aren't yielding like I'd expect. What's the step I'm missing to get my result? Thanks!
type AddBuilder() =
let mutable x = 0
member _.Yield i = x <- x + i
member _.Zero() = 0
member _.Return() = x
let add = AddBuilder()
(* Compiler tells me that each of the numbers in add don't do anything
and suggests putting '|> ignore' in front of each *)
let result = add { 1; 2; 3 }
(* Currently the result is 0 *)
printfn "%i should be 6" result
Note: This is just for creating my own computation expression to expand my learning. Seq.sum would be a better approach. I'm open to the idea that this example completely misses the value of computation expressions and is no good for learning.
There is a lot wrong here.
First, let's start with mere mechanics.
In order for the Yield method to be called, the code inside the curly braces must use the yield keyword:
let result = add { yield 1; yield 2; yield 3 }
But now the compiler will complain that you also need a Combine method. See, the semantics of yield is that each of them produces a finished computation, a resulting value. And therefore, if you want to have more than one, you need some way to "glue" them together. This is what the Combine method does.
Since your computation builder doesn't actually produce any results, but instead mutates its internal variable, the ultimate result of the computation should be the value of that internal variable. So that's what Combine needs to return:
member _.Combine(a, b) = x
But now the compiler complains again: you need a Delay method. Delay is not strictly necessary, but it's required in order to mitigate performance pitfalls. When the computation consists of many "parts" (like in the case of multiple yields), it's often the case that some of them should be discarded. In these situation, it would be inefficient to evaluate all of them and then discard some. So the compiler inserts a call to Delay: it receives a function, which, when called, would evaluate a "part" of the computation, and Delay has the opportunity to put this function in some sort of deferred container, so that later Combine can decide which of those containers to discard and which to evaluate.
In your case, however, since the result of the computation doesn't matter (remember: you're not returning any results, you're just mutating the internal variable), Delay can just execute the function it receives to have it produce the side effects (which are - mutating the variable):
member _.Delay(f) = f ()
And now the computation finally compiles, and behold: its result is 6. This result comes from whatever Combine is returning. Try modifying it like this:
member _.Combine(a, b) = "foo"
Now suddenly the result of your computation becomes "foo".
And now, let's move on to semantics.
The above modifications will let your program compile and even produce expected result. However, I think you misunderstood the whole idea of the computation expressions in the first place.
The builder isn't supposed to have any internal state. Instead, its methods are supposed to manipulate complex values of some sort, some methods creating new values, some modifying existing ones. For example, the seq builder1 manipulates sequences. That's the type of values it handles. Different methods create new sequences (Yield) or transform them in some way (e.g. Combine), and the ultimate result is also a sequence.
In your case, it looks like the values that your builder needs to manipulate are numbers. And the ultimate result would also be a number.
So let's look at the methods' semantics.
The Yield method is supposed to create one of those values that you're manipulating. Since your values are numbers, that's what Yield should return:
member _.Yield x = x
The Combine method, as explained above, is supposed to combine two of such values that got created by different parts of the expression. In your case, since you want the ultimate result to be a sum, that's what Combine should do:
member _.Combine(a, b) = a + b
Finally, the Delay method should just execute the provided function. In your case, since your values are numbers, it doesn't make sense to discard any of them:
member _.Delay(f) = f()
And that's it! With these three methods, you can add numbers:
type AddBuilder() =
member _.Yield x = x
member _.Combine(a, b) = a + b
member _.Delay(f) = f ()
let add = AddBuilder()
let result = add { yield 1; yield 2; yield 3 }
I think numbers are not a very good example for learning about computation expressions, because numbers lack the inner structure that computation expressions are supposed to handle. Try instead creating a maybe builder to manipulate Option<'a> values.
Added bonus - there are already implementations you can find online and use for reference.
1 seq is not actually a computation expression. It predates computation expressions and is treated in a special way by the compiler. But good enough for examples and comparisons.
I'm learning about f# and I understand you don't need to use parentheses when calling a function.
Ex
let addOne arg1 =
arg1 + 1
addOne 1
vs
this.GetType()
Why do I have to use parentheses on the second function?
There is a bit of a mismatch between working with .NET libraries and working with F# libraries when it comes to parameters, but you can generally see () not as parentheses, but as a special value of type unit that means "no useful information".
This means that when you say:
addOne 1
You are calling addOne with a value - number 1 - as a parameter. Now, when you apply the same reading to the second example:
this.GetType()
You can read this as calling this.GetType with a value - the special () unit value as a parameter. If you wanted to be consistent, you could write this with space too:
this.GetType ()
In practice, most people will omit the space when calling .NET libraries. When you do not write the space, F# also supports method chaining so you can write e.g. foo().bar().
Many F# functions taking multiple parameters will use the "curried" form, which means that the parameters need to be separated by spaces. For example:
let add a b = a + b
let mul a b = a * b
add 10 (mul 20 3)
Here, you need parentheses around the second expression, so that the compiler knows how to parse the code. This is in contrast with typical .NET methods, which take parameters as a tuple. F# tuples are written as (10, "hello") and so you can see a method call as an ordinary call accepting a tuple:
some.Operation (10, "Hello")
Again, typically you wouldn't write the space here, because you know this is actually a .NET method call, rather than "passing tuple to a function", but conceptually, you can think of it in both ways.
This is the summary - there are a few corner cases where method calls do not really behave like tuples (e.g. when it comes to named parameters), but this way of thinking about it should give you an idea about how things work.
Are the following two examples equivalent?
Example 1:
let x = String::new();
let y = &x[..];
Example 2:
let x = String::new();
let y = &*x;
Is one more efficient than the other or are they basically the same?
In the case of String and Vec, they do the same thing. In general, however, they aren't quite equivalent.
First, you have to understand Deref. This trait is implemented in cases where a type is logically "wrapping" some lower-level, simpler value. For example, all of the "smart pointer" types (Box, Rc, Arc) implement Deref to give you access to their contents.
It is also implemented for String and Vec: String "derefs" to the simpler str, Vec<T> derefs to the simpler [T].
Writing *s is just manually invoking Deref::deref to turn s into its "simpler form". It is almost always written &*s, however: although the Deref::deref signature says it returns a borrowed pointer (&Target), the compiler inserts a second automatic deref. This is so that, for example, { let x = Box::new(42i32); *x } results in an i32 rather than a &i32.
So &*s is really just shorthand for Deref::deref(&s).
s[..] is syntactic sugar for s.index(RangeFull), implemented by the Index trait. This means to slice the "whole range" of the thing being indexed; for both String and Vec, this gives you a slice of the entire contents. Again, the result is technically a borrowed pointer, but Rust auto-derefs this one as well, so it's also almost always written &s[..].
So what's the difference? Hold that thought; let's talk about Deref chaining.
To take a specific example, because you can view a String as a str, it would be really helpful to have all the methods available on strs automatically available on Strings as well. Rather than inheritance, Rust does this by Deref chaining.
The way it works is that when you ask for a particular method on a value, Rust first looks at the methods defined for that specific type. Let's say it doesn't find the method you asked for; before giving up, Rust will check for a Deref implementation. If it finds one, it invokes it and then tries again.
This means that when you call s.chars() where s is a String, what's actually happening is that you're calling s.deref().chars(), because String doesn't have a method called chars, but str does (scroll up to see that String only gets this method because it implements Deref<Target=str>).
Getting back to the original question, the difference between &*s and &s[..] is in what happens when s is not just String or Vec<T>. Let's take a few examples:
s: String; &*s: &str, &s[..]: &str.
s: &String: &*s: &String, &s[..]: &str.
s: Box<String>: &*s: &String, &s[..]: &str.
s: Box<Rc<&String>>: &*s: &Rc<&String>, &s[..]: &str.
&*s only ever peels away one layer of indirection. &s[..] peels away all of them. This is because none of Box, Rc, &, etc. implement the Index trait, so Deref chaining causes the call to s.index(RangeFull) to chain through all those intermediate layers.
Which one should you use? Whichever you want. Use &*s (or &**s, or &***s) if you want to control exactly how many layers of indirection you want to strip off. Use &s[..] if you want to strip them all off and just get at the innermost representation of the value.
Or, you can do what I do and use &*s because it reads left-to-right, whereas &s[..] reads left-to-right-to-left-again and that annoys me. :)
Addendum
There's the related concept of Deref coercions.
There's also DerefMut and IndexMut which do all of the above, but for &mut instead of &.
They are completely the same for String and Vec.
The [..] syntax results in a call to Index<RangeFull>::index() and it's not just sugar for [0..collection.len()]. The latter would introduce the cost of bound checking. Gladly this is not the case in Rust so they both are equally fast.
Relevant code:
index of String
deref of String
index of Vec (just returns self which triggers the deref coercion thus executes exactly the same code as just deref)
deref of Vec
Suppose the following F# function:
let f (x:int) (y:int) = 42
I suspect that the reason I need to parenthesize the arguments in example z2 below is because of type inference; my example might not be great, but it's easy to imagine how things could get very hairy:
let z1 = f 2 3
let z2 = f 2 (f 3 5)
However, the following case is less clear to me:
let rng = System.Random()
let z3 = f 1 rng.Next(5)
z3 doesn't work, with a clear error message:
error FS0597: Successive arguments should be separated by spaces or
tupled, and arguments involving function or method applications should
be parenthesized.
Fixing it is trivial (parenthesize all the things), but what I am not clear about is why such an expression is a problem. I assume this has to do with type inference again, but naively, it seems to me that here, methods having a list of arguments surrounded by a parenthesis would actually make things less potentially ambiguous. Does this have to do with the fact that rng.Next(5) is equivalent to rng.Next 5?
Can someone hint, give an example or explain why this rule is needed, or what type of problems would arise if it were not there?
I think that the problem here is that the code could be treated as:
let z3 = f 1 rng.Next (5)
This would be equivalent to omitting the parentheses and so it would be calling f with 3 arguments (the second being a function value). This sounds a bit silly, but the compiler actually does not strictly insist on having a space between parameters. For example:
let second a b = b
add 5(1) // This works fine and calls 'add 5 1'
add id(1) // error FS0597
add rng.Next(5) // error FS0597
add (rng.Next(5)) // This works fine (partial application)
I think the problem is that if you look at the sequence of the 4 examples in the above snippet, it is not clear which behavior should you get in the second and the third case.
The call rng.Next(5) is still treated in a special way, because F# allows you to chain calls if they are formed by single-parameter application without space. For example rng.Next(5).ToString(). But, for example, writing second(1)(2) is allowed, but second(1)(2).ToString() will not work.
Could someone please tell me how to write a custom function in Open Office Basic to be used in Open Office Calc and that returns an array of values. An example of one such built-in function is MINVERSE. I need to write a custom function that populates a range of cells in much the same way.
Help would be much appreciated.
Yay, I just figured it out: all you do is return an array from your macro, BUT you also have to press Ctrl+Shift+Enter when typing in the cell formula to call your function (which is also the case when working with other arrays in calc). Here's an example:
Function MakeArray
Dim ret(2,2)
ret(0,0) = 1
ret(1,0) = 2
ret(0,1) = 3
ret(1,1) = 4
MakeArray = ret
End Function
FWIW, damjan's MakeArray function returns a Variant containing an array, I think. (The type returned by MakeArray is unspecified, so it defaults to Variant. A Variant is a container with a descriptive header, apparently cast as needed by the interpreter.)
Almost, but not quite, the same thing as returning an array. According to http://www.cpearson.com/excel/passingandreturningarrays.htm, Microsoft did not introduce the ability to return an array until 2000. His example [ LoadNumbers(Low As Long, High As Long) As Long()] does not compile in OO, flagging a syntax error on the parens following Long. It appears that OO's Basic emulates the pre-2k VBA.