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Return in Scala
I have just started using Scala and have come across some behavior that I don't really understand, and I'm hoping the stackoverflow community can shed some light.
Using this test:
example(1,List(1))
This code works as I would expect - 0 is returned on the second iteration:
def example(value: Int, list: List[Int]): Int = {
if (list.isEmpty)
return 0
else
return 1
example(value - list.head, list.tail) + example(value, list.tail);
}
However, this code does not - it throws a NoSuchElementException("head of empty list") on the second iteration:
def example(value: Int, list: List[Int]): Int = {
if (list.isEmpty)
0
else
1
example(value - list.head, list.tail) + example(value, list.tail);
}
The difference appears to be that in the second example, Scala is assuming "0" and "1" are not return values but expressions to be evaluated because there is another expression at the end of the function. It makes sense that the first example would return as expected due to the explicit "return" keyword.
However, given the syntax of the if statement in second example, I would think the compiler would realize that "0" is not an expression to be evaluated, but rather a return value. Why is this not the case?
You should read the answer here for information about implicit returning since that part of your question is an exact duplicate.
But I'll answer specifically your point about "I would think the compiler would realize that "0" is not an expression to be evaluated, but rather a return value", since it's an interesting observation.
The simple answer is that 0 is an expression... it just happens to be the case that its evaluation is pretty simple. Scala doesn't make a distinction between "expressions that require effort" and "expressions that are easy" when it decides how to process your code. It just processes it the way it's told.
More importantly, if the Scala compiler did get into the business of guessing your intent, that would get pretty crazy, and would make coding a lot more challenging!
To see why we do not want this, let's look at your code that you posted:
def example(value: Int, list: List[Int]): Int = {
if (list.isEmpty)
0
else
1
example(value - list.head, list.tail) + example(value, list.tail);
}
As you mentioned, the 0 and 1 bits aren't really doing anything. So maybe this hypothetical compiler would say "Hey! The only interesting thing I can do with these is to return them!" and thus makes them return. We'll now we've got an if/else in which both sides return. In other words, if list is empty, we return, and if it's not empty, we return. So we always return, and the function ends.
But wait!! There's another line after the if/else expression! So our hypothetical compiler says "Hey! If I return here, then I won't ever execute that line! And that line looks pretty important. Maybe I shouldn't return, so that I can execute it." But then it realizes "Oh no! If i don't return the 0 or 1, then they are pointless! I have to return them!". "But that last line is still there!". "Ahhhhhhh!!!!1!!".
So maybe it's better that the compiler doesn't try to guess what we want and instead just does what we tell it.
Related
I have the following code in Scala:
var i: Int = 0
var sum: Int = 0
for(i<- 1 to 10)
{
sum= sum+1
}
println(sum)
Although it is elementary I noticed that if you write
sum=-sum+1
it produces an error.
Also if you write
sum=-(sum+1)
it produces an error too and
sum= -(sum+1)
produces zero. Why zero?
This doesn't happen in C++ for instance and I found it odd. What is happening and why? Does it matter if you use spaces in Scala?
In the first two cases, the Scala compiler is likely interpreting =- as an operator, instead of associating the - with the right-hand side of the assignment. The third case was answered in the comments.
I was reading the description of the Scala WartRemover tool, and was confused by one of the points they had. The description said this:
return breaks referential transparency. Refactor to terminate
computations in a safe way.
// Won't compile: return is disabled
def foo(n:Int): Int = return n + 1
def foo(ns: List[Int]): Any = ns.map(n => return n + 1)
This doesn't make any sense to me, and both of the examples look referentially transparent. Is there some way in which the return keyword makes it any more likely for a function to break referential transparency? Am I just completely misunderstanding their point?
At it's core, referentially transparency is about evaluating expressions. Fundamentally, it says that if you evaluate an expression in a context, it will evaluate to the same value if you evaluate it in any identical context.
Except that "return" statements don't evaluate to anything at all. They cause the current call of the enclosing method to evaluate to something. There's no way that fits the concept of referential transparency. The "throw" statement has a similar problem.
For the examples, the first one
def foo(n:Int): Int = return n + 1
is benign but verbose and non-idiomatic. The second one
def foo(ns: List[Int]): Any = ns.map(n => return n + 1)
is much more problematic. If passed the empty list, it returns the empty list. If passed a non empty list, it returns the value of the head of the list plus 1.
There is some misunderstanding between me and Scala
0 or 1?
object Fun extends App {
def foo(list:List[Int], count:Int = 0): Int = {
if (list.isEmpty) { // when this is true
return 1 // and we are about to return 1, the code goes to the next line
}
foo(list.tail, count + 1) // I know I do not use "return here" ...
count
}
val result = foo( List(1,2,3) )
println ( result ) // 0
}
Why does it print 0?
Why does recursion work even without "return"
(when it is in the middle of function, but not in the end)?
Why doesn't it return 1? when I use "return" explicitly?
--- EDIT:
It will work if I use return here "return foo(list.tail, count + 1)'.
Bu it does NOT explain (for me) why "return 1" does not work above.
If you read my full explanation below then the answers to your three questions should all be clear, but here's a short, explicit summary for everyone's convenience:
Why does it print 0? This is because the method call was returning count, which had a default value of 0—so it returns 0 and you print 0. If you called it with count=5 then it would print 5 instead. (See the example using println below.)
Why does recursion work even without "return" (when it is in the middle of function, but not in the end)? You're making a recursive call, so the recursion happens, but you weren't returning the result of the recursive call.
Why doesn't it return 1? when I use "return" explicitly? It does, but only in the case when list is empty. If list is non-empty then it returns count instead. (Again, see the example using println below.)
Here's a quote from Programming in Scala by Odersky (the first edition is available online):
The recommended style for methods is in fact to avoid having explicit, and especially multiple, return statements. Instead, think of each method as an expression that yields one value, which is returned. This philosophy will encourage you to make methods quite small, to factor larger methods into multiple smaller ones. On the other hand, design choices depend on the design context, and Scala makes it easy to write methods that have multiple, explicit returns if that's what you desire. [link]
In Scala you very rarely use the return keyword, but instead take advantage that everything in an expression to propagate the return value back up to the top-level expression of the method, and that result is then used as the return value. You can think of return as something more like break or goto, which disrupts the normal control flow and might make your code harder to reason about.
Scala doesn't have statements like Java, but instead everything is an expression, meaning that everything returns a value. That's one of the reasons why Scala has Unit instead of void—because even things that would have been void in Java need to return a value in Scala. Here are a few examples about how expressions work that are relevant to your code:
Things that are expressions in Java act the same in Scala. That means the result of 1+1 is 2, and the result of x.y() is the return value of the method call.
Java has if statements, but Scala has if expressions. This means that the Scala if/else construct acts more like the Java ternary operator. Therefore, if (x) y else z is equivalent to x ? y : z in Java. A lone if like you used is the same as if (x) y else Unit.
A code block in Java is a statement made up of a group of statements, but in Scala it's an expression made up of a group of expressions. A code block's result is the result of the last expression in the block. Therefore, the result of { o.a(); o.b(); o.c() } is whatever o.c() returned. You can make similar constructs with the comma operator in C/C++: (o.a(), o.b(), o.c()). Java doesn't really have anything like this.
The return keyword breaks the normal control flow in an expression, causing the current method to immediately return the given value. You can think of it kind of like throwing an exception, both because it's an exception to the normal control flow, and because (like the throw keyword) the resulting expression has type Nothing. The Nothing type is used to indicate an expression that never returns a value, and thus can essentially be ignored during type inference. Here's simple example showing that return has the result type of Nothing:
def f(x: Int): Int = {
val nothing: Nothing = { return x }
throw new RuntimeException("Can't reach here.")
}
Based on all that, we can look at your method and see what's going on:
def foo(list:List[Int], count:Int = 0): Int = {
// This block (started by the curly brace on the previous line
// is the top-level expression of this method, therefore its result
// will be used as the result/return value of this method.
if (list.isEmpty) {
return 1 // explicit return (yuck)
}
foo(list.tail, count + 1) // recursive call
count // last statement in block is the result
}
Now you should be able to see that count is being used as the result of your method, except in the case when you break the normal control flow by using return. You can see that the return is working because foo(List(), 5) returns 1. In contrast, foo(List(0), 5) returns 5 because it's using the result of the block, count, as the return value. You can see this clearly if you try it:
println(foo(List())) // prints 1 because list is empty
println(foo(List(), 5)) // prints 1 because list is empty
println(foo(List(0))) // prints 0 because count is 0 (default)
println(foo(List(0), 5)) // prints 5 because count is 5
You should restructure your method so that the value that the body is an expression, and the return value is just the result of that expression. It looks like you're trying to write a method that returns the number of items in the list. If that's the case, this is how I'd change it:
def foo(list:List[Int], count:Int = 0): Int = {
if (list.isEmpty) count
else foo(list.tail, count + 1)
}
When written this way, in the base case (list is empty) it returns the current item count, otherwise it returns the result of the recursive call on the list's tail with count+1.
If you really want it to always return 1 you can change it to if (list.isEmpty) 1 instead, and it will always return 1 because the base case will always return 1.
You're returning the value of count from the first call (that is, 0), not the value from the recursive call of foo.
To be more precise, in you code, you don't use the returned value of the recursive call to foo.
Here is how you can fix it:
def foo(list:List[Int], count:Int = 0): Int = {
if (list.isEmpty) {
1
} else {
foo(list.tail, count + 1)
}
}
This way, you get 1.
By the way, don't use return. It doesn't do always what you would expect.
In Scala, a function return implicitly the last value. You don't need to explicitly write return.
Your return works, just not the way you expect because you're ignoring its value. If you were to pass an empty list, you'd get 1 as you expect.
Because you're not passing an empty list, your original code works like this:
foo called with List of 3 elements and count 0 (call this recursion 1)
list is not empty, so we don't get into the block with return
we recursively enter foo, now with 2 elements and count 1 (recursion level 2)
list is not empty, so we don't get into the block with return
we recursively enter foo, now with 1 element and count 2 (recursion level 3)
list is not empty, so we don't get into the block with return
we now enter foo with no elements and count 3 (recursion level 4)
we enter the block with return and return 1
we're back to recursion level 3. The result of the call to foo from which we just came back in neither assigned nor returned, so it's ignored. We proceed to the next line and return count, which is the same value that was passed in, 2
the same thing happens on recursion levels 2 and 1 - we ignore the return value of foo and instead return the original count
the value of count on the recursion level 1 was 0, which is the end result
The fact that you do not have a return in front of foo(list.tail, count + 1) means that, after you return from the recursion, execution is falling through and returning count. Since 0 is passed as a default value for count, once you return from all of the recursed calls, your function is returning the original value of count.
You can see this happening if you add the following println to your code:
def foo(list:List[Int], count:Int = 0): Int = {
if (list.isEmpty) { // when this is true
return 1 // and we are about to return 1, the code goes to the next line
}
foo(list.tail, count + 1) // I know I do not use "return here" ...
println ("returned from foo " + count)
count
}
To fix this you should add a return in front of foo(list.tail.....).
you return count in your program which is a constant and is initialized with 0, so that is what you are returning at the top level of your recursion.
I have the following code, but I can't get it to work. As soon as I place a while loop inside the case, it's returning a unit, no matter what I change within the brackets.
case While(c, body) =>
while (true) {
eval(Num(1))
}
}
How can I make this while loop return a non-Unit type?
I tried adding brackets around my while condition, but still it doesn't do what it's supposed to.
Any pointers?
Update
A little more background information since I didn't really explain what the code should do, which seems to be handy if I want to receive some help;
I have defined a eval(exp : Exp). This will evaluate a function.
Exp is an abstract class. Extended by several classes like Plus, Minus (few more basic operations) and a IfThenElse(cond : Exp, then : Exp, else : Exp). Last but not least, there's the While(cond: Exp, body: Exp).
Example of how it should be used;
eval(Plus(Num(1),Num(4)) would result in NumValue(5). (Evaluation of Num(v : Value) results in NumValue(v). NumValue extends Value, which is another abstract class).
eval(While(Lt(Num(1),Var("n")), Plus(Num(1), Var("n"))))
Lt(a : Exp, b : Exp) returns NumValue(1) if a < b.
It's probably clear from the other answer that Scala while loops always return Unit. What's nice about Scala is that if it doesn't do what you want, you can always extend it.
Here is the definition of a while-like construct that returns the result of the last iteration (it will throw an exception if the loop is never entered):
def whiley[T](cond : =>Boolean)(body : =>T) : T = {
#scala.annotation.tailrec
def loop(previous : T) : T = if(cond) loop(body) else previous
if(cond) loop(body) else throw new Exception("Loop must be entered at least once.")
}
...and you can then use it as a while. (In fact, the #tailrec annotation will make it compile into the exact same thing as a while loop.)
var x = 10
val atExit = whiley(x > 0) {
val squared = x * x
println(x)
x -= 1
squared
}
println("The last time x was printed, its square was : " + atExit)
(Note that I'm not claiming the construct is useful.)
Which iteration would you expect this loop to return? If you want a Seq of the results of all iterations, use a for expression (also called for comprehension). If you want just the last one, create a var outside the loop, set its value on each iteration, and return that var after the loop. (Also look into other looping constructs that are implemented as functions on different types of collections, like foldLeft and foldRight, which have their own interesting behaviors as far as return value goes.) The Scala while loop returns Unit because there's no sensible one size fits all answer to this question.
(By the way, there's no way for the compiler to know this, but the loop you wrote will never return. If the compiler could theoretically be smart enough to figure out that while(true) never terminates, then the expected return type would be Nothing.)
The only purpose of a while loop is to execute a side-effect. Or put another way, it will always evaluate to Unit.
If you want something meaningful back, why don't you consider using an if-else-expression or a for-expression?
As everyone else and their mothers said, while loops do not return values in Scala. What no one seems to have mentioned is that there's a reason for that: performance.
Returning a value has an impact on performance, so the compiler would have to be smart about when you do need that return value, and when you don't. There are cases where that can be trivially done, but there are complex cases as well. The compiler would have to be smarter, which means it would be slower and more complex. The cost was deemed not worth the benefit.
Now, there are two looping constructs in Scala (all the others are based on these two): while loops and recursion. Scala can optimize tail recursion, and the result is often faster than while loops. Or, otherwise, you can use while loops and get the result back through side effects.
False is equivalent to 0 and True is equivalent 1 so it's possible to do something like this:
def bool_to_str(value):
"""value should be a bool"""
return ['No', 'Yes'][value]
bool_to_str(True)
Notice how value is bool but is used as an int.
Is this this kind of use Pythonic or should it be avoided?
I'll be the odd voice out (since all answers are decrying the use of the fact that False == 0 and True == 1, as the language guarantees) as I claim that the use of this fact to simplify your code is perfectly fine.
Historically, logical true/false operations tended to simply use 0 for false and 1 for true; in the course of Python 2.2's life-cycle, Guido noticed that too many modules started with assignments such as false = 0; true = 1 and this produced boilerplate and useless variation (the latter because the capitalization of true and false was all over the place -- some used all-caps, some all-lowercase, some cap-initial) and so introduced the bool subclass of int and its True and False constants.
There was quite some pushback at the time since many of us feared that the new type and constants would be used by Python newbies to restrict the language's abilities, but Guido was adamant that we were just being pessimistic: nobody would ever understand Python so badly, for example, as to avoid the perfectly natural use of False and True as list indices, or in a summation, or other such perfectly clear and useful idioms.
The answers to this thread prove we were right: as we feared, a total misunderstanding of the roles of this type and constants has emerged, and people are avoiding, and, worse!, urging others to avoid, perfectly natural Python constructs in favor of useless gyrations.
Fighting against the tide of such misunderstanding, I urge everybody to use Python as Python, not trying to force it into the mold of other languages whose functionality and preferred style are quite different. In Python, True and False are 99.9% like 1 and 0, differing exclusively in their str(...) (and thereby repr(...)) form -- for every other operation except stringification, just feel free to use them without contortions. That goes for indexing, arithmetic, bit operations, etc, etc, etc.
I'm with Alex. False==0 and True==1, and there's nothing wrong with that.
Still, in Python 2.5 and later I'd write the answer to this particular question using Python's conditional expression:
def bool_to_str(value):
return 'Yes' if value else 'No'
That way there's no requirement that the argument is actually a bool -- just as if x: ... accepts any type for x, the bool_to_str() function should do the right thing when it is passed None, a string, a list, or 3.14.
surely:
def bool_to_str(value):
"value should be a bool"
return 'Yes' if value else 'No'
is more readable.
Your code seems inaccurate in some cases:
>>> def bool_to_str(value):
... """value should be a bool"""
... return ['No', 'Yes'][value]
...
>>> bool_to_str(-2)
'No'
And I recommend you to use just the conditional operator for readability:
def bool_to_str(value):
"""value should be a bool"""
return "Yes" if value else "No"
It is actually a feature of the language that False == 0 and True == 1 (it does not depend on the implementation): Is False == 0 and True == 1 in Python an implementation detail or is it guaranteed by the language?
However, I do agree with most of the other answers: there are more readable ways of obtaining the same result as ['No', 'Yes'][value], through the use of the … if value else … or of a dictionary, which have the respective advantages of hinting and stating that value is a boolean.
Plus, the … if value else … follows the usual convention that non-0 is True: it also works even when value == -2 (value is True), as hinted by dahlia. The list and dict approaches are not as robust, in this case, so I would not recommend them.
Using a bool as an int is quite OK because bool is s subclass of int.
>>> isinstance(True, int)
True
>>> isinstance(False, int)
True
About your code: Putting it in a one-line function like that is over the top. Readers need to find your function source or docs and read it (the name of the function doesn't tell you much). This interrupts the flow. Just put it inline and don't use a list (built at run time), use a tuple (built at compile time if the values are constants). Example:
print foo, bar, num_things, ("OK", "Too many!)[num_things > max_things]
Personally I think it depends on how do you want to use this fact, here are two examples
Just simply use boolean as conditional statement is fine. People do this all the time.
a = 0
if a:
do something
However say you want to count how many items has succeed, the code maybe not very friendly for other people to read.
def succeed(val):
if do_something(val):
return True
else:
return False
count = 0
values = [some values to process]
for val in values:
count += succeed(val)
But I do see the production code look like this.
all_successful = all([succeed(val) for val in values])
at_least_one_successful = any([succeed(val) for val in values])
total_number_of_successful = sum([succeed(val) for val in values])