Converting lambda expression into an expression tree - linq

From C# in depth:
Not all lambda expressions can be converted to expression trees. You
can’t convert a lambda with a block of statements ( even just one
return statement ) into an expresion tree --> it has to be in the form
that just evaluates a single expression.
Since Linq-to-Object statements don't get converted into expression tree objects, the lambda expressions used with Linq-to-Object operators can contain a block of statements
string[] count = { "one ", "two ", "three ", "four ", "five ", "six " };
IEnumerable<int> result = count.Select(item =>
{
Console.WriteLine(item);
return item.Length;
});
foreach (int i in result);
OUTPUT:
one two three four five six
I haven't yet started learning Linq-To-Sql or Linq-To-Entities, but I assume lambda expressions used with LINQ statements that operate on IQueryable<T> can only ever contain a single expression, due to restristion that only a single expression can be converted into an expression tree?
Thank you

It's not just that they can only contain a single expression - it's can't be a statement lambda at all. Even a block like this:
var result = query.Select(item => { return item.Length; });
would be invalid for LINQ to SQL. It would have to be expressed as:
var result = query.Select(item => item.Length);
(I've just noticed that that's the bit of the book you quoted. Oh well - it shows I'm consistent :)
Note that as of .NET 4, expression trees themselves do have the ability to contain blocks, but the C# compiler can't translate statement lambdas into that kind of expression tree. There are other restrictions too, but they rarely cause problems.
This is specified in section 4.6 of the C# 4 spec:
Not all anonymous functions can be represented as expression trees. For instance, anonymous functions with statement bodies and anonymous functions containing assignment expressions cannot be represented. In these cases, a conversion still exists, but will fail at compile-time.

You are correct, they can only ever contain a single expression.

Related

What is the exact difference between the expression and the statement in programming [duplicate]

In Python, what is the difference between expressions and statements?
Expressions only contain identifiers, literals and operators, where operators include arithmetic and boolean operators, the function call operator () the subscription operator [] and similar, and can be reduced to some kind of "value", which can be any Python object. Examples:
3 + 5
map(lambda x: x*x, range(10))
[a.x for a in some_iterable]
yield 7
Statements (see 1, 2), on the other hand, are everything that can make up a line (or several lines) of Python code. Note that expressions are statements as well. Examples:
# all the above expressions
print 42
if x: do_y()
return
a = 7
Expression -- from the New Oxford American Dictionary:
expression: Mathematics a collection
of symbols that jointly express a
quantity : the expression for the
circumference of a circle is 2πr.
In gross general terms: Expressions produce at least one value.
In Python, expressions are covered extensively in the Python Language Reference In general, expressions in Python are composed of a syntactically legal combination of Atoms, Primaries and Operators.
Python expressions from Wikipedia
Examples of expressions:
Literals and syntactically correct combinations with Operators and built-in functions or the call of a user-written functions:
>>> 23
23
>>> 23l
23L
>>> range(4)
[0, 1, 2, 3]
>>> 2L*bin(2)
'0b100b10'
>>> def func(a): # Statement, just part of the example...
... return a*a # Statement...
...
>>> func(3)*4
36
>>> func(5) is func(a=5)
True
Statement from Wikipedia:
In computer programming a statement
can be thought of as the smallest
standalone element of an imperative
programming language. A program is
formed by a sequence of one or more
statements. A statement will have
internal components (e.g.,
expressions).
Python statements from Wikipedia
In gross general terms: Statements Do Something and are often composed of expressions (or other statements)
The Python Language Reference covers Simple Statements and Compound Statements extensively.
The distinction of "Statements do something" and "expressions produce a value" distinction can become blurry however:
List Comprehensions are considered "Expressions" but they have looping constructs and therfore also Do Something.
The if is usually a statement, such as if x<0: x=0 but you can also have a conditional expression like x=0 if x<0 else 1 that are expressions. In other languages, like C, this form is called an operator like this x=x<0?0:1;
You can write you own Expressions by writing a function. def func(a): return a*a is an expression when used but made up of statements when defined.
An expression that returns None is a procedure in Python: def proc(): pass Syntactically, you can use proc() as an expression, but that is probably a bug...
Python is a bit more strict than say C is on the differences between an Expression and Statement. In C, any expression is a legal statement. You can have func(x=2); Is that an Expression or Statement? (Answer: Expression used as a Statement with a side-effect.) The assignment statement of x=2 inside of the function call of func(x=2) in Python sets the named argument a to 2 only in the call to func and is more limited than the C example.
Though this isn't related to Python:
An expression evaluates to a value.
A statement does something.
>>> x + 2 # an expression
>>> x = 1 # a statement
>>> y = x + 1 # a statement
>>> print y # a statement (in 2.x)
2
An expression is something that can be reduced to a value, for example "1+3" is an expression, but "foo = 1+3" is not.
It's easy to check:
print(foo = 1+3)
If it doesn't work, it's a statement, if it does, it's an expression.
Another statement could be:
class Foo(Bar): pass
as it cannot be reduced to a value.
Statements represent an action or command e.g print statements, assignment statements.
print 'hello', x = 1
Expression is a combination of variables, operations and values that yields a result value.
5 * 5 # yields 25
Lastly, expression statements
print 5*5
An expression is something, while a statement does something.
An expression is a statement as well, but it must have a return.
>>> 2 * 2         #expression
>>> print(2 * 2)     #statement
PS:The interpreter always prints out the values of all expressions.
An expression is a statement that returns a value. So if it can appear on the right side of an assignment, or as a parameter to a method call, it is an expression.
Some code can be both an expression or a statement, depending on the context. The language may have a means to differentiate between the two when they are ambiguous.
STATEMENT:
A Statement is a action or a command that does something. Ex: If-Else,Loops..etc
val a: Int = 5
If(a>5) print("Hey!") else print("Hi!")
EXPRESSION:
A Expression is a combination of values, operators and literals which yields something.
val a: Int = 5 + 5 #yields 10
Expressions always evaluate to a value, statements don't.
e.g.
variable declaration and assignment are statements because they do not return a value
const list = [1,2,3];
Here we have two operands - a variable 'sum' on the left and an expression on the right.
The whole thing is a statement, but the bit on the right is an expression as that piece of code returns a value.
const sum = list.reduce((a, b)=> a+ b, 0);
Function calls, arithmetic and boolean operations are good examples of expressions.
Expressions are often part of a statement.
The distinction between the two is often required to indicate whether we require a pice of code to return a value.
References
Expressions and statements
2.3 Expressions and statements - thinkpython2 by Allen B. Downey
2.10. Statements and Expressions - How to Think like a Computer Scientist by Paul Resnick and Brad Miller
An expression is a combination of values, variables, and operators. A value all by itself is
considered an expression, and so is a variable, so the following are all legal expressions:
>>> 42
42
>>> n
17
>>> n + 25
42
When you type an expression at the prompt, the interpreter evaluates it, which means that
it finds the value of the expression. In this example, n has the value 17 and n + 25 has the
value 42.
A statement is a unit of code that has an effect, like creating a variable or displaying a
value.
>>> n = 17
>>> print(n)
The first line is an assignment statement that gives a value to n. The second line is a print
statement that displays the value of n.
When you type a statement, the interpreter executes it, which means that it does whatever
the statement says. In general, statements don’t have values.
An expression translates to a value.
A statement consumes a value* to produce a result**.
*That includes an empty value, like: print() or pop().
**This result can be any action that changes something; e.g. changes the memory ( x = 1) or changes something on the screen ( print("x") ).
A few notes:
Since a statement can return a result, it can be part of an expression.
An expression can be part of another expression.
Statements before could change the state of our Python program: create or update variables, define function, etc.
And expressions just return some value can't change the global state or local state in a function.
But now we got :=, it's an alien!
Expressions:
Expressions are formed by combining objects and operators.
An expression has a value, which has a type.
Syntax for a simple expression:<object><operator><object>
2.0 + 3 is an expression which evaluates to 5.0 and has a type float associated with it.
Statements
Statements are composed of expression(s). It can span multiple lines.
A statement contains a keyword.
An expression does not contain a keyword.
print "hello" is statement, because print is a keyword.
"hello" is an expression, but list compression is against this.
The following is an expression statement, and it is true without list comprehension:
(x*2 for x in range(10))
Python calls expressions "expression statements", so the question is perhaps not fully formed.
A statement consists of pretty much anything you can do in Python: calculating a value, assigning a value, deleting a variable, printing a value, returning from a function, raising an exception, etc. The full list is here: http://docs.python.org/reference/simple_stmts.html#
An expression statement is limited to calling functions (e.g.,
math.cos(theta)"), operators ( e.g., "2+3"), etc. to produce a value.

Explain the below Linq Query?

results.Where(x=>x.Members.Any(y=>members.Contains(y.Name.ToLower())
I happened to see this query in internet. Can anyone explain this query please.
suggest me a good LINQ tutorial for this newbie.
thank you all.
Edited:
what is this x and y stands for?
x is a single result, of the type of the elements in the results sequence.
y is a single member, of the type of the elements in the x.Members sequence.
These are lambda expressions (x => x.whatever) that were introduced into the language with C# 3, where x is the input, and the right side (x.whatever) is the output (in this particular usage scenario).
An easier example
var list = new List<int> { 1, 2, 3 };
var oddNumbers = list.Where(i => i % 2 != 0);
Here, i is a single int item that is an input into the expression. i % 2 != 0 is a boolean expression evaluating whether the input is even or odd. The entire expression (i => i % 2 != 0) is a predicate, a Func<int, bool>, where the input is an integer and the output is a boolean. Follow? As you iterate over the query oddNumbers, each element in the list sequence is evaluated against the predicate. Those that pass then become part of your output.
foreach (var item in oddNumbers)
Console.WriteLine(item);
// writes 1, 3
Its a lambda expression. Here is a great LINQ tutorial
Interesting query, but I don't like it.
I'll answer your second question first. x and y are parameters to the lambda methods that are defined in the calls to Where() and Any(). You could easy change the names to be more meaningful:
results.Where(result =>
result.Members.Any(member => members.Contains(member.Name.ToLower());
And to answer your first question, this query will return each item in results where the Members collection has at least one item that is also contained in the Members collection as a lower case string.
The logic there doesn't make a whole lot of sense to me with knowing what the Members collection is or what it holds.
x will be every instance of the results collection. The query uses lambda syntax, so x=>x.somemember means "invoke somemember on each x passed in. Where is an extension method for IEnumerables that expects a function that will take an argument and return a boolean. Lambda syntax creates delegates under the covers, but is far more expressive for carrying out certain types of operation (and saves a lot of typing).
Without knowing the type of objects held in the results collection (results will be something that implements IEnumerable), it is hard to know exactly what the code above will do. But an educated guess is that it will check all the members of all the x's in the above collection, and return you an IEnumerable of only those that have members with all lower-case names.

Using Where() with a dynamic Func fails, but works with a hard coded Where() clause

See the two Linq (to SharePoint) code samples below.
The only differences are the highlighted sections of code. The first statement works as expected with a hard-coded where clause, but the 2nd set of code throws the error “Value does not fall in within the expected range” when I try to do a count on the items. What am I missing?
Works
relatedListItems = dc.GetList<GeneralPage>("Pages")
.Where(x => x.RelatedPracticesTitle.Any(y=>y=="Foo"))
if (relatedListItems.Count() == 0)
{…}
Fails - “Value does not fall within the expected range”
Func<GeneralPage, bool> f = x => x.RelatedPracticesTitle.Any(y => y == "Foo");
relatedListItems = dc.GetList<GeneralPage>("Pages")
.Where(f)
if (relatedListItems.Count() == 0)
{…}
If it's LINQ to Sharepoint, presumably that means it should be using expression trees, not delegates. Try:
Expression<Func<GeneralPage, bool>> f =
x => x.RelatedPracticesTitle.Any(y => y == "Foo");
relatedListItems = dc.GetList<GeneralPage>("Pages").Where(f);
By the way, it's generally a better idea to use Any() rather than Count() if you just want to find out if there are any results - that way it can return as soon as it's found the first one. (It also expresses what you're interested in more clearly, IMO.)
In the first case, you're using the Expression<Func<GeneralPage, bool>> overload and pass an expression which I assume LINQ to SharePoint will try to convert to CAML and execute.
In the second case, you're passing the plain Func<GeneralPage, bool> so LINQ to SharePoint can't figure out how to compose a query (it only sees the delegate, not the expression).

What are the precise rules for when you can omit parenthesis, dots, braces, = (functions), etc.?

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.

Are curly brackets used in Lua?

If curly brackets ('{' and '}') are used in Lua, what are they used for?
Table literals.
The table is the central type in Lua, and can be treated as either an associative array (hash table or dictionary) or as an ordinary array. The keys can be values of any Lua type except nil, and the elements of a table can hold any value except nil.
Array member access is made more efficient than hash key access behind the scenes, but the details don't usually matter. That actually makes handling sparse arrays handy since storage only need be allocated for those cells that contain a value at all.
This does lead to a universal 1-based array idiom that feels a little strange to a C programmer.
For example
a = { 1, 2, 3 }
creates an array stored in the variable a with three elements that (coincidentally) have the same values as their indices. Because the elements are stored at sequential indices beginning with 1, the length of a (given by #a or table.getn(a)) is 3.
Initializing a table with non-integer keys can be done like this:
b = { one=1, pi=3.14, ["half pi"]=1.57, [function() return 17 end]=42 }
where b will have entries named "one", "pi", "half pi", and an anonymous function. Of course, looking up that last element without iterating the table might be tricky unless a copy of that very function is stored in some other variable.
Another place that curly braces appear is really the same semantic meaning, but it is concealed (for a new user of Lua) behind some syntactic sugar. It is common to write functions that take a single argument that should be a table. In that case, calling the function does not require use of parenthesis. This results in code that seems to contain a mix of () and {} both apparently used as a function call operator.
btn = iup.button{title="ok"}
is equivalent to
btn = iup.button({title="ok"})
but is also less hard on the eyes. Incidentally, calling a single-argument function with a literal value also works for string literals.
list/ditionary constructor (i.e. table type constructor).
They are not used for code blocks if that's what you mean. For that Lua just uses the end keyword to end the block.
See here
They're used for table literals as you would use in C :
t = {'a', 'b', 'c'}
That's the only common case. They're not used for block delimiters. In a lua table, you can put values of different types :
t={"foo", 'b', 3}
You can also use them as dictionnaries, à la Python :
t={name="foo", age=32}

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