I have a few test cases that I need to write a method for, and how they execute the method is like this:
Planets with: aString , anotherString
Where "Planets" is a class and "with:" is a class method. But my confusion is in the way that arguments are given, because I was under the impression that methods with multiple arguments get executed like "methodName: arg1 Arg2: arg2", but in here the two arguments are separated by comma. Can somebody explain this to me please?
It has to do with the order of precedence for messages. Unary messages get handled first, then binary messages, and finally keyword messages.
Unary messages have no arguments (new, sqrt, isPrime)
Binary messages go between two objects (like a + or in your case the ,).
Keyword messages end in a colon (like with:) and sometimes there is more than one (like to: do:)
The binary message , (concatenate two strings) is processed first because it has a higher precedence. Then the keyword message with: is processed after because it has a lower precedence. Messages inside of parentheses automatically get the highest precedence and are evaluated first. When there is a "tie" (multiple messages have the same precedence), they are evaluated left to right.
One weird thing about Smalltalk is that + - * / all have the same binary message precedence. So 8 + 2 * 4 evaluates to 40 not 16 like you might expect.
Most likely, the class method #with: receives a String as its only argument. This means that #with: is a keyword message with arity 1. In the expression
Planets with: aString , anotherString
there are two messages: (1) the message with selector #with: and (2) the message with selector #,. The latter has a receiver aString and an argument anotherString. This is a binary message with arity 1 too.
In fact, one could equally write
Planets with: (aString , anotherString)
to emphasize the role of argument played by the parenthesized expression. We usually don't do that because precedence rules make it unnecessary.
Now note that in the declaration of the method with selector#with: you would have
Planets class >> #with: aString
"some code here"
But one thing is the declaration of a method and another its invocation, in your case
Plantes with: aString , anotherString
is an expression aimed at sending the message #with: to Planets with an argument that happens to be the concatenation of two strings.
In sum, when you declare/define the method the argument is aString, but when you invoke it (i.e., when you send the message) the argument is anything that evaluates to a String.
As a side note I would indicate that Planets isn't a good name for a class. In fact, class names are usually in singular, as they represent a concept, not a collection of instances of said concept.
foo = 1
p +foo
This example code prints 1 just like if the + was not there. I know - in front of a variable gets the opposite of what the variable was (-29 becomes 29) but is there any case where a variable with a + in front of it ever does anything or can I safely remove it every time I see it? To clarify this a bit I am asking no specifically about numbers assigned to variables but any datatype in ruby.
+ is both a unary operator (one argument) and a binary operator (two arguments). It is defined on the Numeric class. Unary operators are defined using # suffix to differentiate from the binary operator.
Unary Plus—Returns the receiver.
This is the source for the method:
num_uplus(VALUE num)
{
return num;
}
So to answer your question,
Does + in front of a variable in ruby ever do anything?
NO, for Numeric values.
I just looked it up for strings and yes it does do something for frozen strings.
If the string is frozen, then return duplicated mutable string.
If the string is not frozen, then return the string itself.
static VALUE
str_uplus(VALUE str)
{
if (OBJ_FROZEN(str)) {
return rb_str_dup(str);
}
else {
return str;
}
}
is there any case where a variable with a + in front of it ever does anything
Yes. Every time. It calls the +# method.
or can I safely remove it every time I see it?
No, you can't. It will change the semantics of your code: before, it will call the +# method, after, it won't.
Whether or not that changes the outcome of your program, depends on what that method is doing. The default implementation for Numeric#+# simply returns self, but of course someone could have monkey-patched it to do something different.
Also, String#+# does something more interesting: if self is a frozen string, it will return an unfrozen, mutable copy; if self is already mutable, it returns self.
Other objects in the core library don't have a +# method, so they will usually raise a NoMethodError. If you remove the call, they won't; that is also a behavioral change.
This prints 1:
def sum(i)
i=i+[2]
end
$x=[1]
sum($x)
print $x
This prints 12:
def sum(i)
i.push(2)
end
$x=[1]
sum($x)
print $x
The latter is modifying the global variable $x. Why is it modified in the second example and not in the first one? Will this will happen with any method (not only push) of the class Array?
Variable scope is irrelevant here.
In the first code, you are only assigning to a variable i using the assignment operator =, whereas in the second code, you are modifying $x (also referred to as i) using a destructive method push. Assignment never modifies any object. It just provides a name to refer to an object. Methods are either destructive or non-destructive. Destructive methods like Array#push, String#concat modify the receiver object. Non-destructive methods like Array#+, String#+ do not modify the receiver object, but create a new object and return that, or return an already existing object.
Answer to your comment
Whether or not you can modify the receiver depends on the class of the receiver object. For arrays, hashes, and strings, etc., which are said to be mutable, it is possible to modify the receiver. For numerals, etc, which are said to be immutable, it is impossible to do that.
In the first snippet, you assign new local variable to hold result of $x + [2] operation which is returned, but it doesn't change $x (because + method doesn't modify receiver object). In your second snipped, you use Array#push method, which modifies an object (in this case, object assigned to $x global var and passed as i into your sum method) on which it's called.
i.push(2) appends 2 to the array pointed by i. Since this is the same array pointed by $x, $x gets 2 appended to it as well.
i=i+[2] creates a new array and set i to it - and now this is a different array than the one pointed by $x.
I am looking at some old VB6 code. I am new to VB and I come from a C/Java background, so I don't understand some of the assignment statements. Here is one example -
Private Type UGH
Rsp(3) As Byte
ProgramId(7) As Byte
RID(7) As Byte
TID(3) As Byte
FL(39) As Byte
End Type
Private UHeader As UGH
Later, the assignment takes place as follows-
With UHeader
StringToByteArray UHeader.ProgramId(), "ABCDPQRS"
My question is, why is the parenthesis used after ProgramId in the above assignment? To me it seems like a function call, but it obviously is not a function call. Then what is it?
Well it is not actually needed. Calling the same line without the parenthesis will do exactly the same thing. The reason it is allowed is because it is a byte array.
Maybe rewriting the line like this will make it more readable:
call StringToByteArray(UHeader.ProgramId(), "ABCDPQRS")
But this is also valid:
call StringToByteArray(UHeader.ProgramId, "ABCDPQRS")
It would probably make more sense to you if the StringToByteArray method was a function instead of a subroutine:
Private Function StringToByteArray(ByVal strValue As String) As Byte()
'conversion code left out
End Function
Then you can call it like this:
UHeader.ProgramId() = StringToByteArray("ABCDPQRS")
or this:
UHeader.ProgramId = StringToByteArray("ABCDPQRS")
The parenthesis in this case is used to designate an array. For Rsp(3) as Byte, it indicates an array of 3 bytes indexed from 0 to 2.
For the line: StringToByteArray UHeader.ProgramId(), "ABCDPQRS", StringToByteArray is a method that takes a byte array--when passing an array to a method, you pass it in as array().
Parenthesis are also used for array syntax.
And so it's said, StringToByteArray UHeader.ProgramId(), "ABCDPQRS" is not strictly an assignment, it is a method (subroutine or function) call. StringToByteArray is going to be defined somewhere in your program and it will probably take a ByRef array parameter, a string parameter, and it probably copys the string parameter into the the byte buffer.
UHeader.ProgramId is actually a pointer to the array. You can simply do this: StringToByteArray UHeader.ProgramID, "ABCDPQRS". The rule in VB6 is that Functions require parenthesization of argument lists and Subs require non-parenthesization, unless preceded by the Call statement in which case they do.
Since you are passing the array by reference (the default), operations performed in the subroutine are performed on the array itself. Basically, you are passing a pointer to an array and a literal string, and the subroutine will populate the array with the characters in the string. Sort of like an "informal" function call, if you will.
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.