What are the precise rules for when you can omit (omit) parentheses, dots, braces, = (functions), etc.?
For example,
(service.findAllPresentations.get.first.votes.size) must be equalTo(2).
service is my object
def findAllPresentations: Option[List[Presentation]]
votes returns List[Vote]
must and be are both functions of specs
Why can't I go:
(service findAllPresentations get first votes size) must be equalTo(2)
?
The compiler error is:
"RestServicesSpecTest.this.service.findAllPresentations
of type
Option[List[com.sharca.Presentation]]
does not take parameters"
Why does it think I'm trying to pass in a parameter? Why must I use dots for every method call?
Why must (service.findAllPresentations get first votes size) be equalTo(2) result in:
"not found: value first"
Yet, the "must be equalTo 2" of
(service.findAllPresentations.get.first.votes.size) must be equalTo 2, that is, method chaining works fine? - object chain chain chain param.
I've looked through the Scala book and website and can't really find a comprehensive explanation.
Is it in fact, as Rob H explains in Stack Overflow question Which characters can I omit in Scala?, that the only valid use-case for omitting the '.' is for "operand operator operand" style operations, and not for method chaining?
You seem to have stumbled upon the answer. Anyway, I'll try to make it clear.
You can omit dot when using the prefix, infix and postfix notations -- the so called operator notation. While using the operator notation, and only then, you can omit the parenthesis if there is less than two parameters passed to the method.
Now, the operator notation is a notation for method-call, which means it can't be used in the absence of the object which is being called.
I'll briefly detail the notations.
Prefix:
Only ~, !, + and - can be used in prefix notation. This is the notation you are using when you write !flag or val liability = -debt.
Infix:
That's the notation where the method appears between an object and it's parameters. The arithmetic operators all fit here.
Postfix (also suffix):
That notation is used when the method follows an object and receives no parameters. For example, you can write list tail, and that's postfix notation.
You can chain infix notation calls without problem, as long as no method is curried. For example, I like to use the following style:
(list
filter (...)
map (...)
mkString ", "
)
That's the same thing as:
list filter (...) map (...) mkString ", "
Now, why am I using parenthesis here, if filter and map take a single parameter? It's because I'm passing anonymous functions to them. I can't mix anonymous functions definitions with infix style because I need a boundary for the end of my anonymous function. Also, the parameter definition of the anonymous function might be interpreted as the last parameter to the infix method.
You can use infix with multiple parameters:
string substring (start, end) map (_ toInt) mkString ("<", ", ", ">")
Curried functions are hard to use with infix notation. The folding functions are a clear example of that:
(0 /: list) ((cnt, string) => cnt + string.size)
(list foldLeft 0) ((cnt, string) => cnt + string.size)
You need to use parenthesis outside the infix call. I'm not sure the exact rules at play here.
Now, let's talk about postfix. Postfix can be hard to use, because it can never be used anywhere except the end of an expression. For example, you can't do the following:
list tail map (...)
Because tail does not appear at the end of the expression. You can't do this either:
list tail length
You could use infix notation by using parenthesis to mark end of expressions:
(list tail) map (...)
(list tail) length
Note that postfix notation is discouraged because it may be unsafe.
I hope this has cleared all the doubts. If not, just drop a comment and I'll see what I can do to improve it.
Class definitions:
val or var can be omitted from class parameters which will make the parameter private.
Adding var or val will cause it to be public (that is, method accessors and mutators are generated).
{} can be omitted if the class has no body, that is,
class EmptyClass
Class instantiation:
Generic parameters can be omitted if they can be inferred by the compiler. However note, if your types don't match, then the type parameter is always infered so that it matches. So without specifying the type, you may not get what you expect - that is, given
class D[T](val x:T, val y:T);
This will give you a type error (Int found, expected String)
var zz = new D[String]("Hi1", 1) // type error
Whereas this works fine:
var z = new D("Hi1", 1)
== D{def x: Any; def y: Any}
Because the type parameter, T, is inferred as the least common supertype of the two - Any.
Function definitions:
= can be dropped if the function returns Unit (nothing).
{} for the function body can be dropped if the function is a single statement, but only if the statement returns a value (you need the = sign), that is,
def returnAString = "Hi!"
but this doesn't work:
def returnAString "Hi!" // Compile error - '=' expected but string literal found."
The return type of the function can be omitted if it can be inferred (a recursive method must have its return type specified).
() can be dropped if the function doesn't take any arguments, that is,
def endOfString {
return "myDog".substring(2,1)
}
which by convention is reserved for methods which have no side effects - more on that later.
() isn't actually dropped per se when defining a pass by name paramenter, but it is actually a quite semantically different notation, that is,
def myOp(passByNameString: => String)
Says myOp takes a pass-by-name parameter, which results in a String (that is, it can be a code block which returns a string) as opposed to function parameters,
def myOp(functionParam: () => String)
which says myOp takes a function which has zero parameters and returns a String.
(Mind you, pass-by-name parameters get compiled into functions; it just makes the syntax nicer.)
() can be dropped in the function parameter definition if the function only takes one argument, for example:
def myOp2(passByNameString:(Int) => String) { .. } // - You can drop the ()
def myOp2(passByNameString:Int => String) { .. }
But if it takes more than one argument, you must include the ():
def myOp2(passByNameString:(Int, String) => String) { .. }
Statements:
. can be dropped to use operator notation, which can only be used for infix operators (operators of methods that take arguments). See Daniel's answer for more information.
. can also be dropped for postfix functions
list tail
() can be dropped for postfix operators
list.tail
() cannot be used with methods defined as:
def aMethod = "hi!" // Missing () on method definition
aMethod // Works
aMethod() // Compile error when calling method
Because this notation is reserved by convention for methods that have no side effects, like List#tail (that is, the invocation of a function with no side effects means that the function has no observable effect, except for its return value).
() can be dropped for operator notation when passing in a single argument
() may be required to use postfix operators which aren't at the end of a statement
() may be required to designate nested statements, ends of anonymous functions or for operators which take more than one parameter
When calling a function which takes a function, you cannot omit the () from the inner function definition, for example:
def myOp3(paramFunc0:() => String) {
println(paramFunc0)
}
myOp3(() => "myop3") // Works
myOp3(=> "myop3") // Doesn't work
When calling a function that takes a by-name parameter, you cannot specify the argument as a parameter-less anonymous function. For example, given:
def myOp2(passByNameString:Int => String) {
println(passByNameString)
}
You must call it as:
myOp("myop3")
or
myOp({
val source = sourceProvider.source
val p = myObject.findNameFromSource(source)
p
})
but not:
myOp(() => "myop3") // Doesn't work
IMO, overuse of dropping return types can be harmful for code to be re-used. Just look at specification for a good example of reduced readability due to lack of explicit information in the code. The number of levels of indirection to actually figure out what the type of a variable is can be nuts. Hopefully better tools can avert this problem and keep our code concise.
(OK, in the quest to compile a more complete, concise answer (if I've missed anything, or gotten something wrong/inaccurate please comment), I have added to the beginning of the answer. Please note this isn't a language specification, so I'm not trying to make it exactly academically correct - just more like a reference card.)
A collection of quotes giving insight into the various conditions...
Personally, I thought there'd be more in the specification. I'm sure there must be, I'm just not searching for the right words...
There are a couple of sources however, and I've collected them together, but nothing really complete / comprehensive / understandable / that explains the above problems to me...:
"If a method body has more than one
expression, you must surround it with
curly braces {…}. You can omit the
braces if the method body has just one
expression."
From chapter 2, "Type Less, Do More", of Programming Scala:
"The body of the upper method comes
after the equals sign ‘=’. Why an
equals sign? Why not just curly braces
{…}, like in Java? Because semicolons,
function return types, method
arguments lists, and even the curly
braces are sometimes omitted, using an
equals sign prevents several possible
parsing ambiguities. Using an equals
sign also reminds us that even
functions are values in Scala, which
is consistent with Scala’s support of
functional programming, described in
more detail in Chapter 8, Functional
Programming in Scala."
From chapter 1, "Zero to Sixty: Introducing Scala", of Programming Scala:
"A function with no parameters can be
declared without parentheses, in which
case it must be called with no
parentheses. This provides support for
the Uniform Access Principle, such
that the caller does not know if the
symbol is a variable or a function
with no parameters.
The function body is preceded by "="
if it returns a value (i.e. the return
type is something other than Unit),
but the return type and the "=" can be
omitted when the type is Unit (i.e. it
looks like a procedure as opposed to a
function).
Braces around the body are not
required (if the body is a single
expression); more precisely, the body
of a function is just an expression,
and any expression with multiple parts
must be enclosed in braces (an
expression with one part may
optionally be enclosed in braces)."
"Functions with zero or one argument
can be called without the dot and
parentheses. But any expression can
have parentheses around it, so you can
omit the dot and still use
parentheses.
And since you can use braces anywhere
you can use parentheses, you can omit
the dot and put in braces, which can
contain multiple statements.
Functions with no arguments can be
called without the parentheses. For
example, the length() function on
String can be invoked as "abc".length
rather than "abc".length(). If the
function is a Scala function defined
without parentheses, then the function
must be called without parentheses.
By convention, functions with no
arguments that have side effects, such
as println, are called with
parentheses; those without side
effects are called without
parentheses."
From blog post Scala Syntax Primer:
"A procedure definition is a function
definition where the result type and
the equals sign are omitted; its
defining expression must be a block.
E.g., def f (ps) {stats} is
equivalent to def f (ps): Unit =
{stats}.
Example 4.6.3 Here is a declaration
and a de?nition of a procedure named
write:
trait Writer {
def write(str: String)
}
object Terminal extends Writer {
def write(str: String) { System.out.println(str) }
}
The code above is implicitly completed
to the following code:
trait Writer {
def write(str: String): Unit
}
object Terminal extends Writer {
def write(str: String): Unit = { System.out.println(str) }
}"
From the language specification:
"With methods which only take a single
parameter, Scala allows the developer
to replace the . with a space and omit
the parentheses, enabling the operator
syntax shown in our insertion operator
example. This syntax is used in other
places in the Scala API, such as
constructing Range instances:
val firstTen:Range = 0 to 9
Here again, to(Int) is a vanilla
method declared inside a class
(there’s actually some more implicit
type conversions here, but you get the
drift)."
From Scala for Java Refugees Part 6: Getting Over Java:
"Now, when you try "m 0", Scala
discards it being a unary operator, on
the grounds of not being a valid one
(~, !, - and +). It finds that "m" is
a valid object -- it is a function,
not a method, and all functions are
objects.
As "0" is not a valid Scala
identifier, it cannot be neither an
infix nor a postfix operator.
Therefore, Scala complains that it
expected ";" -- which would separate
two (almost) valid expressions: "m"
and "0". If you inserted it, then it
would complain that m requires either
an argument, or, failing that, a "_"
to turn it into a partially applied
function."
"I believe the operator syntax style
works only when you've got an explicit
object on the left-hand side. The
syntax is intended to let you express
"operand operator operand" style
operations in a natural way."
Which characters can I omit in Scala?
But what also confuses me is this quote:
"There needs to be an object to
receive a method call. For instance,
you cannot do “println “Hello World!”"
as the println needs an object
recipient. You can do “Console
println “Hello World!”" which
satisfies the need."
Because as far as I can see, there is an object to receive the call...
I find it easier to follow this rule of thumb: in expressions spaces alternate between methods and parameters. In your example, (service.findAllPresentations.get.first.votes.size) must be equalTo(2) parses as (service.findAllPresentations.get.first.votes.size).must(be)(equalTo(2)). Note that the parentheses around the 2 have a higher associativity than the spaces. Dots also have higher associativity, so (service.findAllPresentations.get.first.votes.size) must be.equalTo(2)would parse as (service.findAllPresentations.get.first.votes.size).must(be.equalTo(2)).
service findAllPresentations get first votes size must be equalTo 2 parses as service.findAllPresentations(get).first(votes).size(must).be(equalTo).2.
Actually, on second reading, maybe this is the key:
With methods which only take a single
parameter, Scala allows the developer
to replace the . with a space and omit
the parentheses
As mentioned on the blog post: http://www.codecommit.com/blog/scala/scala-for-java-refugees-part-6 .
So perhaps this is actually a very strict "syntax sugar" which only works where you are effectively calling a method, on an object, which takes one parameter. e.g.
1 + 2
1.+(2)
And nothing else.
This would explain my examples in the question.
But as I said, if someone could point out to be exactly where in the language spec this is specified, would be great appreciated.
Ok, some nice fellow (paulp_ from #scala) has pointed out where in the language spec this information is:
6.12.3:
Precedence and associativity of
operators determine the grouping of
parts of an expression as follows.
If there are several infix operations in an expression, then
operators with higher precedence bind
more closely than operators with lower
precedence.
If there are consecutive infix operations e0 op1 e1 op2 . . .opn en
with operators op1, . . . , opn of the
same precedence, then all these
operators must have the same
associativity. If all operators are
left-associative, the sequence is
interpreted as (. . . (e0 op1 e1) op2
. . .) opn en. Otherwise, if all
operators are rightassociative, the
sequence is interpreted as e0 op1 (e1
op2 (. . .opn en) . . .).
Postfix operators always have lower precedence than infix operators. E.g.
e1 op1 e2 op2 is always equivalent to
(e1 op1 e2) op2.
The right-hand operand of a
left-associative operator may consist
of several arguments enclosed in
parentheses, e.g. e op (e1, . . .
,en). This expression is then
interpreted as e.op(e1, . . . ,en).
A left-associative binary operation e1
op e2 is interpreted as e1.op(e2). If
op is rightassociative, the same
operation is interpreted as { val
x=e1; e2.op(x ) }, where x is a fresh
name.
Hmm - to me it doesn't mesh with what I'm seeing or I just don't understand it ;)
There aren't any. You will likely receive advice around whether or not the function has side-effects. This is bogus. The correction is to not use side-effects to the reasonable extent permitted by Scala. To the extent that it cannot, then all bets are off. All bets. Using parentheses is an element of the set "all" and is superfluous. It does not provide any value once all bets are off.
This advice is essentially an attempt at an effect system that fails (not to be confused with: is less useful than other effect systems).
Try not to side-effect. After that, accept that all bets are off. Hiding behind a de facto syntactic notation for an effect system can and does, only cause harm.
I'm trying to use a ternary expression along with a continue or break construct inside a do-while loop in Dart but I get a compilation error.
do
{
expr ? print('Conditionally print something') : continue;
}while(expr == true);
The above code fails at compile-time but if I use a pair of if-else decision structure, the code works. So, the question to the community is why the ternary expression is not working along with continue or break construct?
The ternary operator takes 3 expressions in the form of (expr1) ? (expr2) : (expr3).
You can't execute statements in ternary operator, nor only in dart but in other languages also. Since break and continue are not expressions but statements they cant be used here
No, you can't do it.
continue is a Dart Statement (read below), and print is a function
First
In the Dart Language Specification section 17.23 explains how it works.
https://dart.dev/guides/language/specifications/DartLangSpec-v2.10.pdf
Search the original document, because the copy/paste doesn't seems to work well.
17.23 Conditional conditional
A conditional expression evaluates one of two expressions based on a boolean
condition.
{conditionalExpression} ::= {ifNullExpression}
(‘?’ {expressionWithoutCascade} ‘:’ {expressionWithoutCascade})?
Evaluation of a conditional expression c of the form e1?e2 : e3 proceeds as
follows:
First, e1 is evaluated to an object o1. It is a dynamic error if the runtime type of o1 is not bool. If r is true, then the value of c is the result of
evaluating the expression e2. Otherwise the value of c is the result of evaluating
the expression e3.
Second
As you can see the ternary operator requires expressions, but continue, in the same PDF of language specification is defined as an statement, a reserved word, as:
18 Statements statements
A statement is a fragment of Dart code that can be executed at run time.
Statements, unlike expressions, do not evaluate to an object, but are instead
executed for their effect on the program state and control flow
Third
in the case of print, it's taken as a function I guess, didn't find the specification. Perhaps it returns void.
We can still ask ourselves, why can't we put continue in a function, even in a lambda like () => { continue; } or similar. The short answer is that as said in the specification for the continue statement, if it's not inside a while, etc. is gonna give a compile error. And if it's inside a function, it will prevent that function to reach the return statement, and again, the ternary will expect a return value.
We can still research a little more, many things are inside the specification.
When something like this happens, you can also search without specifying the language, to get information on JAVA or C# that may help you.
Java: Ternary with no return. (For method calling)
It seems that Swift language would allow continue and break inside the ternary, as per this article - https://forums.swift.org/t/bringing-control-flow-keywords-to-the-ternary-operator/13878
Because a ternary operator is returning a value. With the following syntax:
condition ? expresion1 : expression2
which means:
If condition is true, it return expression1. If it is not it return expression2.
So, you can't use statement like this:
print('Conditionally print something')
or
continue
An expression evaluates to a value. A statement does something.
In Javascript you can use the spread syntax in a function call like this:
console.log(...[1,2,3]);
Is there an equivalent in Reason? I tried the following:
let bound = (number, lower, upper) => {
max(lower, min(upper, number));
};
let parameters = (1,0,20);
bound(...parameters) |> Js.log;
But this gives an unknown syntax error:
Try reason snippet
There's not. Reason is a statically typed language, and lists are dynamically-sized and homogenous. It would be of very limited use, and not at all obvious how it would deal with too few or too many arguments. If you want to pass a list, you should just accept a list and deal with it appropriately, as a separate function if desired.
You could of course use a tuple instead, which is fixed-size and heterogenous, but I don't see a use-case for that either, since you might as well just call the function directly then.
For JavaScript FFI there is however the bs.splice attribute, which will allow you to apply a variable number of arguments to a js function using an array. But it needs to be called with an array literal, not just any array.
I created a sample project and a framework next to it. The framework is called "SampleFramework". Then I created a custom operator in SampleFramework. Here is what it looks like:
infix operator >>= {associativity left}
public func >>=<A, B>(a: A?, f: A -> B?) -> B? {
if let a = a { return f(a) }
else { return .None }
}
Then I wanted to use it my main application. I imported the SampleFramework to my source file and then I wrote this code to test it:
NSURL(string: "www.google.com") >>= { println("\($0)") }
It didn't compile. Here is Xcode's error message:
Ambiguous operator declarations found for operator. Operator is not a
known binary operator
I can confirm that that what you are seeing is what really happens. I just tried it myself and I've seen the same result.
My opinion is that >>= is somehow conflicting with some other operator (probably the bite shift operator: >>) or is being declared somewhere else too (you can see why I think that here). I did successfully declared custom operators in a framework and used those in the main app code before (you can see that here for example).
What I would suggest is rename your custom operator to something else. When I did that (renamed the custom operator to >>>=) the compiler stopped complaining and my app compiled just fine.
Later edit
Ok. So this might help a bit more. Basically when an operator is already declared and you want to add extra functionality to that operator (for example doing things like 3 * "Hello" like Johan Kool said he wanted to) all you have to do is overload that operator's method.
Basically, in your specific case I am now 100% that >>= is an already declared operator and you can go ahead and just add these lines in your framework:
public func >>=<A, B>(a: A?, f: A -> B?) -> B? {
if let a = a { return f(a) }
else { return .None }
}
This will make your operator work. BUT it will inherit the precedence and associativity of the original operator thus giving you less control over how it's supposed to behave.
I figured it out. I think declaration of operator (infix operator >>= {associativity left}) is module specific. You can't apply access control to it. But the operator function can be accessed in other modules. So in order to make this work I had to copy operator declaration and paste it to main project.
It seems to be sufficient to just add the
infix operator + : Additive
in each source where you want to apply your framework-defined operator, assuming that your framework has some
public func + (lhs: ...
declared.
Anyhow, it's nasty to have the need for adding this if you use a framework and I'd like to see some way to have some global approach where importing a framework also makes available the operators automatically.
Edit After fiddling around for hours, the problem went away. I have not figured out what's the cause. I had lots of infix declarations in my basic framework (for historic reason). Once I got rid of them, everything went to normal. So now that I have only infix for the new operators, it's fine. But obviously re-declaring existing ones (like *, +, etc.) Swift seems to get hick-ups.
If we have a method:
def hello() = "world"
I'm told that we can call it as:
hello()
also
hello
They both work and will output world, but why?
PS:
I see some words in this https://stackoverflow.com/a/12340289/342235:
No, actually, they are not. Even though they both call a method without parameters, one is a "method with zero parameter lists" while the other is a "method with one empty parameter list"
But I still not understand why hello would work
Scala allows the omission of parentheses on methods of arity-0 (no arguments):
You should only omit the parenthesis when there are no side effects to the invokation though
http://docs.scala-lang.org/style/method-invocation.html
As stated in Oliver Shaw's answer, Scala lets you leave out the parenthesis in functions with 0 arguments. As for why, it's likely to facilitate easy refactoring.
If you have a function that takes no arguments, and produces no side-effects, it's equivalent to an immutable value. If you always call such functions without parentheses, then you're free to change the underlying definition of the function to a value without having to refactor it everywhere it's referenced.
It's worth noting that val definitions are actually modeled as 0-arity methods in scala (unless specified with a private final beforehand).
Scala does something similar with its treatment of arity-1 functions defined on classes. You are allowed to omit the dot and the parenthesis. For example:
case class Foo(value: String) {
def prepend(value2: String) = Foo(value2 + value)
}
Foo("one").prepend("two")
// is the same as...
Foo("one") prepend "two"
This is because Scala models all operators as arity-1 functions. You can rewrite 1 + 2 as 1.+(2) and have it mean the same thing. Representing operators as fully-fledged functions has some nice qualities. You can expect to pass an operator anywhere that you could pass a function, and the definition of the operator is actually defined for class instances (as opposed to a language like C#, where the infix operators are actually static methods that use special syntactic sugar to let them be represented as infix).