+= operator in Scala - scala

I'm reading Programming in Scala by M. Odersky and now I'm trying to understand the meaning of operators. As far as I can see, any operator in Scala is just a method. Consider the following example:
class OperatorTest(var a : Int) {
def +(ot: OperatorTest): OperatorTest = {
val retVal = OperatorTest(0);
retVal.a = a + ot.a;
println("=")
return retVal;
}
}
object OperatorTest {
def apply(a: Int) = new OperatorTest(a);
}
I this case we have only + operator defined in this class. And if we type something like this:
var ot = OperatorTest(10);
var ot2 = OperatorTest(20);
ot += ot2;
println(ot.a);
then
=+
30
will be the output. So I'd assume that for each class (or type?) in Scala we have += operator defined for it, as a += b iff a = a + b. But since every operator is just a method, where the += operator defined? Maybe there is some class (like Object in Java) containing all the defenitions for such operators and so forth.
I looked at AnyRef in hoping to find, but couldn't.

+= and similar operators are desugared by the compiler in case there is a + defined and no += is defined. (Similarly works for other operators too.) Check the Scala Language Specification (6.12.4):
Assignment operators are treated specially in that they can be
expanded to assignments if no other interpretation is valid.
Let's consider an assignment operator such as += in an infix operation
l += r, where l, r are expressions. This operation can be
re-interpreted as an operation which corresponds to the assignment
l = l + r except that the operation's left-hand-side l is evaluated
only once.
The re-interpretation occurs if the following two conditions are
fulfilled.
The left-hand-side l does not have a member named +=, and also cannot
be converted by an implicit conversion to a value with a member named
+=. The assignment l = l + r is type-correct. In particular this implies that l refers to a variable or object that can be assigned to,
and that is convertible to a value with a member named +.

Related

When does a += b become a = a + b in Scala?

Scala only sometimes desugars
a += b
to
a = a + b
but not always. For example, some mutable collections define a += method, where instead it becomes
a.+=(b)
Is this behaviour
entirely determined by whether there is a suitable += method on a? (incl. are there any other examples of this behaviour?)
independent of whether the objects are val or var?
Relevant example
Adapted from Programming in Scala, in
var s = Set("a", "b")
s += "c"
In this case, the second line of code s += "c" is essentially shorthand for:
s = s + "c"
When does a += b become a = a + b in Scala?
When there is no applicable += method, there is an applicable + method and a is assignable (that is, it's a var or there's an a_= method).
Or as the spec puts it:
The re-interpretation occurs if the following two conditions are fulfilled.
The left-hand-side l does not have a member named +=, and also cannot be converted by an implicit conversion to a value with a member named +=.
The assignment l = l + r is type-correct. In particular this implies that l refers to a variable or object that can be assigned to, and that is convertible to a value with a member named +.
Is this behaviour
entirely determined by whether there is a suitable += method on a?
independent of whether the objects are val or var?
Not quite. If there is a suitable += method, it will be called regardless of any other factors (such as a being assignable). But when there isn't, the other factors determine whether it's desugared or you get an error message.
Note that the error message you get is different than the one you'd get from the desugared version: When the criteria for the desugaring don't apply, you get an error message that tells you "+= is not a member of ...", plus an explanation why the desugaring couldn't be applied (such as "receiver is not assignable" or the type error you'd get from a + b if a + b would produce a type error).
In my first answer I was a little hasty, my apologies. After researching a little and after read better the #sepp2k's answer and comment, I came to the conclusion that some classes in Scala implements the += method, and other just the + method, I played a little in With some Scala code and for instance:
//Set, Int, Double, String implements the "+" method,
//and then "+=" is syntactic sugar of a = a + b
var set = Set("a", "b")
set += "c"
var num = 3
num += 2
var str = "43"
str += 5
var l = List()
l += "someString"
// As you mention, MutableList implements "+=" method, and when you do
// mutL += 4, is the same as call the method mutL.+=
var mutL = new mutable.MutableList[Int]
mutL += 4

Where is Scala's += defined in the context of Int?

Just starting out with Scala
var c = 0
c += 1 works
c.+= gives me error: value += is not a member of Int
Where is the += defined?
Section 6.12.4 Assignment Operators of the Scala Language Specification (SLS) explains how such compound assignment operators are desugared:
l ω= r
(where ω is any sequence of operator characters other than <, >, ! and doesn't start with =) gets desugared to
l.ω=(r)
IFF l has a member named ω= or is implicitly convertible to an object that has a member named ω=.
Otherwise, it gets desugared to
l = l.ω(r)
(except l is guaranteed to be only evaluated once), if that typechecks.
Or, to put it more simply: the compiler will first try l.ω=(r) and if that doesn't work, it will try l = l.ω(r).
This allows something like += to work like it does in other languages but still be overridden to do something different.
Actually, the code you've described does work.
scala> var c = 4
c: Int = 4
scala> c.+=(2) // no output because assignment is not an expression
scala> c
res1: Int = 6
I suspect (but can't say for sure) that it can't be found in the library because the compiler de-surgars (rewrites) it to c = c.+(1), which is in the library.

Where is the `++=` operator defined for a scala immutable list?

How is the ++= being evaluated in the code below? The ++= operator is being applied to a class that is explicitly defined as an immutable list, which does not have the ++= operator. x ++= y appears to have the same behavior as x = x ++ y, but I cannot find where this behavior is coming form. What is happening here? (and where is it documented?)
object ListConcatenation extends App {
var list = scala.collection.immutable.List[Int](1)
println(list)
list = list ++ scala.collection.immutable.List[Int](2)
println(list)
// This operator does not exist on the List class, but has the same behavior as the operation above.
list ++= scala.collection.immutable.List[Int](3)
println(list)
}
Output
List(1)
List(1, 2)
List(1, 2, 3)
x ++= y appears to have the same behavior as x = x ++ y
It does.
but I cannot find where this behavior is coming form.
It's built into the language. When x op= y can't be interpreted as a calling the method op= (because no such method exists in x's class or any class that it can be implicitly converted to), it's instead interpreted as calling the method op and an assignment.
(and where is it documented?)
In Scala's language specification, section 6.12.4:
Assignment Operators
An assignment operator is an operator symbol (syntax category op in Identifiers) that ends in an equals character “=”, with the exception of operators for which one of the following conditions holds:
the operator also starts with an equals character, or
the operator is one of (<=), (>=), (!=).
Assignment operators are treated specially in that they can be expanded to assignments if no other interpretation is valid.
Let's consider an assignment operator such as += in an infix operation l += r, where l, r are expressions. This operation can be re-interpreted as an operation which corresponds to the assignment
l = l + r
except that the operation's left-hand-side l is evaluated only once.
The re-interpretation occurs if the following two conditions are fulfilled.
The left-hand-side l does not have a member named +=, and also cannot be converted by an implicit conversion to a value with a member named +=.
The assignment l = l + r is type-correct. In particular this implies that l refers to a variable or object that can be assigned to, and that is convertible to a value with a member named +.

Where is += method located for int in scala

Is +=(or any assignment operators) a method in scala for Int type.
For example,
var x=5
x+=1
Here I am able to use += method only when it is a variable.
I am not able to do,
5+=1
Does scala compiler considers this method as a special case?
Why it is not available in scala.Int class?
There is no += method, it is expanded to x = x + 1 by the compiler. This is detailed in the specification:
6.12.4 Assignment Operators
Let's consider an assignment operator such as += in an infix operation
l += r, where l, r
are expressions. This operation can be re-interpreted as an operation
which corresponds to the assignment
l = l + r
except that the operation's left-hand-side l is evaluated only once.
The re-interpretation occurs if the following two conditions are
fulfilled.
The left-hand-side l does not have a member named +=, and also cannot be converted by an implicit conversion to a value with a member
named +=.
The assignment l = l + r is type-correct. In particular this implies that l refers to a variable or object that can be assigned to,
and that is convertible to a value with a member named +.

Scala puts precedence on implicit conversion over "natural" operations... Why? Is this a bug? Or am I doing something wrong?

This simple test, of course, works as expected:
scala> var b = 2
b: Int = 2
scala> b += 1
scala> b
res3: Int = 3
Now I bring this into scope:
class A(var x: Int) { def +=(y:Int) { this.x += y } }
implicit def int2A(i:Int) : A = new A(i)
I'm defining a new class and a += operation on it, and a convenient implicit conversion for those times when I want to add an Int to A's Int value.
I never expected this would affect the way my regular Int operations behave, when the "A" class is not at all part of the expression.
But it does:
scala> var b:Int = 0
b: Int = 0
scala> b += 1
scala> b
res29: Int = 0
scala> b += 2
scala> b
res31: Int = 0
What seems to be happening here is that the b:Int is implicitly converted to an "A", which is not bound to any variable, and then += is invoked on it, discarding the results.
Scala seems to give high precedence the implicit conversion over the natural += behavior (compiler magic, not an actual method) that is already defined to Ints. Common-sense as well as a C++ background tells me implicits should only be invoked as a last resort, when the compilation would otherwise fail. That leads to several questions...
Why? Is this a bug? Is it by design?
Is there a work-around (other than not using "+=" for my DSL's "+=" operation)?
Thanks
As others have noted, Int cannot have a += "method", because Int is immutable. What happens instead is that x += 1 is treated as a short form for x = x + 1, but only if there is no
method called += that is defined on the type. So method resolution takes precedence.
Given that Scala lets you define += methods and also lets you do += on variables, could we have changed the priority of the two? I.e. try expanded += first and only if that fails search for a method named +=?
Theoretically yes, but I argue it would have been worse than the current scheme. Practically, no. There are many types in Scala's collection library that define both a + method for
non-destructive addition and a += method for destructive addition. If we had switched the priority around then a call like
myHashTable += elem
would expand to
myHashTable = myHashTable + elem
So it would construct a new hashtable and assign this back to the variable, instead of simply updating an element. Not a wise thing to do...
From Programming in Scala, Chapter 17:
Whenever you write a += b, and a does
not support a method named +=, Scala
will try interpreting it as a = a + b.
The class Int does not contain method +=. However class A provides += method. That might be triggering the implicit conversion from Int to A.
I don't think it is a bug.
Actually, Int only has a "+" method but doesn't have a "+=" method.
b += 1 would transform to b = b + 1 in compile time if there is not a other implicit which has a "+=" method exists.
Scala seems to give high precedence the implicit conversion over the natural += that is already defined to Ints.
Which version of Scala are you talking about? I don't know of any version that has a += method on Int. Certainly, none of the still supported versions do, that must be some really ancient version you have there.
And since there is no += on Int, but you are calling a += method on Int, Scala tries to satisfy that type constraint via an implicit conversion.
Even withstanding Eastsun's explanation, it seems like this is a bug, and it should try the b=b+1 transformation before trying an implicit conversion for +=.
Please ask this question to the scala-user email list by emailing scala-user#listes.epfl.ch or by visiting n4.nabble.com/Scala-User-f1934582.html. If it's a bug, that's where it will be noticed and fixed.