Related
I am passing Logical Foundations course and became stuck upon the last excersize of Basics:
Having binary number write a converter to it's unary representation:
Inductive bin : Type :=
| Z
| A (n : bin)
| B (n : bin).
Fixpoint bin_to_nat (m:bin) : nat :=
(* What to do here? *)
I solved the problem with a recursive function in C. The only thing, I used "0" istead of "A" and "1" instead of "B".
#include <stdio.h>
unsigned int pow2(unsigned int power)
{
if(power != 0)
return 2 << (power - 1);
else
return 1;
}
void rec_converter(char str[], size_t i)
{
if(str[i] == 'Z')
printf("%c", 'Z');
else if(str[i] == '0')
rec_converter(str, ++i);
else if(str[i] == '1')
{
unsigned int n = pow2(i);
for (size_t j = 0; j < n; j++)
{
printf("%c", 'S');
}
rec_converter(str, ++i);
}
}
int main(void)
{
char str[] = "11Z";
rec_converter(str, 0);
printf("\n");
return 0;
}
My problem now is how to write this code in coq:
unsigned int n = pow2(i);
for (size_t j = 0; j < n; j++)
{
printf("%c", 'S');
}
rec_converter(str, ++i);
The main difference between your code and the Coq code is that the Coq code ought to return the natural number, rather than printing it. That means we'll need to keep track of everything that your solution printed and return the result all at once.
Since printing an S means that the answer is the successor of whatever else is printed, we'll need a function that can take the 2^(n)th successor of a natural number. There are various ways to do this, but I'd suggest recursion on n and noting that the 2^(n + 1)th successor of x is the 2^(n)th successor of the 2^(n)th successor of x.
That should be enough to get what you want.
unsigned int n = pow2(i);
for (size_t j = 0; j < n; j++)
{
printf("%c", 'S');
}
rec_converter(str, ++i);
can be written (in pseudo-Coq) as
pow2_succ i (rec_converter str (S i)).
However, one other thing to note: you may not be able to directly access the ith "character" of the input, but this shouldn't be a problem. When you write your function as a Fixpoint
Fixpoint rec_converter (n: bin) (i: nat): nat :=
match n with
| Z => 0
| A m => ...
| B m => ...
end.
the first "character" of m will be the second "character" of the original input. So you'll just need to access the first "character", which is exactly what a Fixpoint does.
For the question on computing powers of 2, you should look at the following file, provided in the Coq libraries (at least up to version 8.9):
https://coq.inria.fr/distrib/current/stdlib/Coq.Init.Nat.html
This file contains a host of functions around the natural numbers, they could all be used as illustrations about how to program with Coq and this datatype.
Fixpoint bin_to_nat (m:bin) : nat :=
match m with
| Z => O
| A n =>2 * (bin_to_nat n)
| B n =>2 * (bin_to_nat n) + 1
end.
see: coq art's 2004. P167-P168. ( How to understand 'positive' type in Coq)
In programming languages like Scala or Lua, we can define nested functions such as
function factorial(n)
function _fac(n, acc)
if n == 0 then
return acc
else
return _fac(n-1, acc * n)
end
end
return _fac(n, 1)
end
Does this approach cause any inefficiency because the nested function instance is defined, or created, everytime we invoke the outer function?
Does this approach cause any inefficiency because the nested function
instance is defined, or created, everytime we invoke the outer
function?
Efficiency is a large and broad topic. I am assuming that by "inefficient" you mean "does calling the method recursively each time have an overhead"?
I can only speak on Scala's behalf, specifically the flavor targeting the JVM, as other flavors may act differently.
We can divide this question into two, depending on what you really meant.
Nested (local scope) methods in Scala are a lexical scope feature, meaning they give you the accessibility to outer method values, but once we emit the bytecode, they are defined at the class level, just as a plain java method.
For completeness, do know that Scala also has function values, which are first class citizens, meaning you can pass them around to other functions, then these would incur an allocation overhead, since they are implemented using classes.
Factorial can be written in a tail recursive manner, as you wrote it in your example. The Scala compiler is intelligent enough such that it will notice your method is tail recursive and turn it into an iterative loop, avoiding the method invocation for each iteration. It may also, if found possible, attempt to inline the factorial method, avoiding the overhead of an additional stack frame allocation.
For example, consider the following factorial implementation in Scala:
def factorial(num: Int): Int = {
#tailrec
def fact(num: Int, acc: Int): Int = num match {
case 0 => acc
case n => fact(n - 1, acc * n)
}
fact(num, 1)
}
On the face of it, we have a recursive method. Let's see what the JVM bytecode looks like:
private final int fact$1(int, int);
Code:
0: iload_1
1: istore 4
3: iload 4
5: tableswitch { // 0 to 0
0: 24
default: 28
}
24: iload_2
25: goto 41
28: iload 4
30: iconst_1
31: isub
32: iload_2
33: iload 4
35: imul
36: istore_2
37: istore_1
38: goto 0
41: ireturn
What we see here is that the recursion turned into an iterative loop (a tableswitch + a jump instruction).
Regarding the method instance creation, if our method was not tail recursive, the JVM runtime would need to interpret it for each invocation, until the C2 compiler finds it sufficient such that it will JIT compile it and re-use the machine code for each method call afterwards.
Generally, I would say this shouldn't worry you unless you've noticed the method is on the execution of your hot path and profiling the code led you to ask this question.
To conclude, efficiency is a very delicate, use case specific topic. I think we don't have enough information to tell you, from the simplified example you've provided, if this is the best option to choose for your use case. I say again, if this isn't something that showed up on your profiler, don't worry about this.
Let's benchmark it in Lua with/without nested functions.
Variant 1 (inner function object is created on every call)
local function factorial1(n)
local function _fac1(n, acc)
if n == 0 then
return acc
else
return _fac1(n-1, acc * n)
end
end
return _fac1(n, 1)
end
Variant 2 (functions are not nested)
local function _fac2(n, acc)
if n == 0 then
return acc
else
return _fac2(n-1, acc * n)
end
end
local function factorial2(n)
return _fac2(n, 1)
end
Benchmarking code (calculate 12! 10 mln times and display used CPU time in seconds):
local N = 1e7
local start_time = os.clock()
for j = 1, N do
factorial1(12)
end
print("CPU time of factorial1 = ", os.clock() - start_time)
local start_time = os.clock()
for j = 1, N do
factorial2(12)
end
print("CPU time of factorial2 = ", os.clock() - start_time)
Output for Lua 5.3 (interpreter)
CPU time of factorial1 = 8.237
CPU time of factorial2 = 6.074
Output for LuaJIT (JIT-compiler)
CPU time of factorial1 = 1.493
CPU time of factorial2 = 0.141
The answer depends on the language of course.
What happens in Scala in particular is that inner functions are compiled as they were standing outside of the scope of the function within which they are defined.
In this way the language only allows you to invoke them from the lexical scope where they are defined in, but does not actually instantiate the function multiple times.
We can easily test this by compiling two variants of the
The first one is a fairly faithful port of your Lua code:
class Function1 {
def factorial(n: Int): Int = {
def _fac(n: Int, acc: Int): Int =
if (n == 0)
acc
else
_fac(n-1, acc * n)
_fac(n, 1)
}
}
The second one is more or less the same, but the tail recursive function is defined outside of the scope of factorial:
class Function2 {
def factorial(n: Int): Int = _fac(n, 1)
private final def _fac(n: Int, acc: Int): Int =
if (n == 0)
acc
else
_fac(n-1, acc * n)
}
We can now compile these two classes with scalac and then use javap to have a look at the compiler output:
javap -p Function*.scala
which will yield the following output
Compiled from "Function1.scala"
public class Function1 {
public int factorial(int);
private final int _fac$1(int, int);
public Function1();
}
Compiled from "Function2.scala"
public class Function2 {
public int factorial(int);
private final int _fac(int, int);
public Function2();
}
I added the private final keywords to minimize the difference between the two, but the main thing to notice is that in both cases the definitions appear at the class level, with inner functions automatically defined as private and final and with a small decoration to ensure no name class (e.g. if you define a loop inner function inside two different ones).
Not sure about Lua or other languages, but I can expect at least most compiled languages to adopt a similar approach.
Yes (or it used to), as evidenced by Lua's effort to reuse function values when execution passes through a function definition multiple times.
Lua 5.2 Changes
Equality between function values has changed. Now, a function
definition may not create a new value; it may reuse some previous
value if there is no observable difference to the new function.
Since you have coded (assuming Lua) a function assigned to a global or local declared at a higher scope, you could code the short-circuit yourself (presuming no other code sets it to anything other than nil or false):
function factorial(n)
_fac = _fac or function (n, acc)
…
end
…
end
I don't know about lua, but in Scala are very common and used in recursive functions to ensure tail-safe optimization:
def factorial(i: Int): Int = {
#tailrec
def fact(i: Int, accumulator: Int): Int = {
if (i <= 1)
accumulator
else
fact(i - 1, i * accumulator)
}
fact(i, 1)
}
More info about tail-safe and recursion here
I'm trying to write some code as below -
def kthSmallest(matrix: Array[Array[Int]], k: Int): Int = {
val pq = new PriorityQueue[Int]() //natural ordering
var count = 0
for (
i <- matrix.indices;
j <- matrix(0).indices
) yield {
pq += matrix(i)(j)
count += 1
} //This would yield Any!
pq.dequeue() //kth smallest.
}
My question is, that I only want to loop till the time count is less than k (something like takeWhile(count != k)), but as I'm also inserting elements into the priority queue in the yield, this won't work in the current state.
My other options are to write a nested loop and return once count reaches k. Is it possible to do with yield? I could not find a idiomatic way of doing it yet. Any pointers would be helpful.
It's not idiomatic for Scala to use vars or break loops. You can go for recursion, lazy evaluation or duct tape a break, giving up on some performance (just like return, it's implemented as an Exception, and won't perform well enough). Here are the options broken down:
Use recursion - recursive algorithms are the analog of loops in functional languages
def kthSmallest(matrix: Array[Array[Int]], k: Int): Int = {
val pq = new PriorityQueue[Int]() //natural ordering
#tailrec
def fillQueue(i: Int, j: Int, count: Int): Unit =
if (count >= k || i >= matrix.length) ()
else {
pq += matrix(i)(j)
fillQueue(
if (j >= matrix(i).length - 1) i + 1 else i,
if (j >= matrix(i).length - 1) 0 else j + 1,
count + 1)
}
fillQueue(0, 0, 0)
pq.dequeue() //kth smallest.
}
Use a lazy structure, as chengpohi suggested - this doesn't sound very much like a pure function though. I'd suggest to use an Iterator instead of a Stream in this case though - as iterators don't memoize the steps they've gone through (might spare some memory for large matrices).
For those desperately willing to use break, Scala supports it in an attachable fashion (note the performance caveat mentioned above):
import scala.util.control.Breaks
breakable {
// loop code
break
}
There is a way using the Stream lazy evaluation to do this. Since for yield is equal to flatMap, you can convert for yield to flatMap with Stream:
matrix.indices.toStream.flatMap(i => {
matrix(0).indices.toStream.map(j => {
pq += matrix(i)(j)
count += 1
})
}).takeWhile(_ => count <= k)
Use toStream to convert the collection to Stream, and Since Stream is lazy evaluation, so we can use takeWhile to predicate count to terminate the less loops without init others.
Recently I had an interview for Scala Developer position. I was asked such question
// matrix 100x100 (content unimportant)
val matrix = Seq.tabulate(100, 100) { case (x, y) => x + y }
// A
for {
row <- matrix
elem <- row
} print(elem)
// B
val func = print _
for {
row <- matrix
elem <- row
} func(elem)
and the question was: Which implementation, A or B, is more efficent?
We all know that for comprehensions can be translated to
// A
matrix.foreach(row => row.foreach(elem => print(elem)))
// B
matrix.foreach(row => row.foreach(func))
B can be written as matrix.foreach(row => row.foreach(print _))
Supposedly correct answer is B, because A will create function print 100 times more.
I have checked Language Specification but still fail to understand the answer. Can somebody explain this to me?
In short:
Example A is faster in theory, in practice you shouldn't be able to measure any difference though.
Long answer:
As you already found out
for {xs <- xxs; x <- xs} f(x)
is translated to
xxs.foreach(xs => xs.foreach(x => f(x)))
This is explained in §6.19 SLS:
A for loop
for ( p <- e; p' <- e' ... ) e''
where ... is a (possibly empty) sequence of generators, definitions, or guards, is translated to
e .foreach { case p => for ( p' <- e' ... ) e'' }
Now when one writes a function literal, one gets a new instance every time the function needs to be called (§6.23 SLS). This means that
xs.foreach(x => f(x))
is equivalent to
xs.foreach(new scala.Function1 { def apply(x: T) = f(x)})
When you introduce a local function type
val g = f _; xxs.foreach(xs => xs.foreach(x => g(x)))
you are not introducing an optimization because you still pass a function literal to foreach. In fact the code is slower because the inner foreach is translated to
xs.foreach(new scala.Function1 { def apply(x: T) = g.apply(x) })
where an additional call to the apply method of g happens. Though, you can optimize when you write
val g = f _; xxs.foreach(xs => xs.foreach(g))
because the inner foreach now is translated to
xs.foreach(g())
which means that the function g itself is passed to foreach.
This would mean that B is faster in theory, because no anonymous function needs to be created each time the body of the for comprehension is executed. However, the optimization mentioned above (that the function is directly passed to foreach) is not applied on for comprehensions, because as the spec says the translation includes the creation of function literals, therefore there are always unnecessary function objects created (here I must say that the compiler could optimize that as well, but it doesn't because optimization of for comprehensions is difficult and does still not happen in 2.11). All in all it means that A is more efficient but B would be more efficient if it is written without a for comprehension (and no function literal is created for the innermost function).
Nevertheless, all of these rules can only be applied in theory, because in practice there is the backend of scalac and the JVM itself which both can do optimizations - not to mention optimizations that are done by the CPU. Furthermore your example contains a syscall that is executed on every iteration - it is probably the most expensive operation here that outweighs everything else.
I'd agree with sschaef and say that A is the more efficient option.
Looking at the generated class files we get the following anonymous functions and their apply methods:
MethodA:
anonfun$2 -- row => row.foreach(new anonfun$2$$anonfun$1)
anonfun$2$$anonfun$1 -- elem => print(elem)
i.e. matrix.foreach(row => row.foreach(elem => print(elem)))
MethodB:
anonfun$3 -- x => print(x)
anonfun$4 -- row => row.foreach(new anonfun$4$$anonfun$2)
anonfun$4$$anonfun$2 -- elem => func(elem)
i.e. matrix.foreach(row => row.foreach(elem => func(elem)))
where func is just another indirection before calling to print. In addition func needs to be looked up, i.e. through a method call on an instance (this.func()) for each row.
So for Method B, 1 extra object is created (func) and there are # of elem additional function calls.
The most efficient option would be
matrix.foreach(row => row.foreach(func))
as this has the least number of objects created and does exactly as you would expect.
Benchmark
Summary
Method A is nearly 30% faster than method B.
Link to code: https://gist.github.com/ziggystar/490f693bc39d1396ef8d
Implementation Details
I added method C (two while loops) and D (fold, sum). I also increased the size of the matrix and used an IndexedSeq instead. Also I replaced the print with something less heavy (sum all entries).
Strangely the while construct is not the fastest. But if one uses Array instead of IndexedSeq it becomes the fastest by a large margin (factor 5, no boxing anymore). Using explicitly boxed integers, methods A, B, C are all equally fast. In particular they are faster by 50% compared to the implicitly boxed versions of A, B.
Results
A
4.907797735
4.369745787
4.375195012000001
4.7421321800000005
4.35150636
B
5.955951859000001
5.925475619
5.939570085000001
5.955592247
5.939672226000001
C
5.991946029
5.960122757000001
5.970733164
6.025532582
6.04999499
D
9.278486201
9.265983922
9.228320372
9.255641645
9.22281905
verify results
999000000
999000000
999000000
999000000
>$ scala -version
Scala code runner version 2.11.0 -- Copyright 2002-2013, LAMP/EPFL
Code excerpt
val matrix = IndexedSeq.tabulate(1000, 1000) { case (x, y) => x + y }
def variantA(): Int = {
var r = 0
for {
row <- matrix
elem <- row
}{
r += elem
}
r
}
def variantB(): Int = {
var r = 0
val f = (x:Int) => r += x
for {
row <- matrix
elem <- row
} f(elem)
r
}
def variantC(): Int = {
var r = 0
var i1 = 0
while(i1 < matrix.size){
var i2 = 0
val row = matrix(i1)
while(i2 < row.size){
r += row(i2)
i2 += 1
}
i1 += 1
}
r
}
def variantD(): Int = matrix.foldLeft(0)(_ + _.sum)
How do I break out a loop?
var largest=0
for(i<-999 to 1 by -1) {
for (j<-i to 1 by -1) {
val product=i*j
if (largest>product)
// I want to break out here
else
if(product.toString.equals(product.toString.reverse))
largest=largest max product
}
}
How do I turn nested for loops into tail recursion?
From Scala Talk at FOSDEM 2009 http://www.slideshare.net/Odersky/fosdem-2009-1013261
on the 22nd page:
Break and continue
Scala does not have them. Why?
They are a bit imperative; better use many smaller functions
Issue how to interact with closures.
They are not needed!
What is the explanation?
You have three (or so) options to break out of loops.
Suppose you want to sum numbers until the total is greater than 1000. You try
var sum = 0
for (i <- 0 to 1000) sum += i
except you want to stop when (sum > 1000).
What to do? There are several options.
(1a) Use some construct that includes a conditional that you test.
var sum = 0
(0 to 1000).iterator.takeWhile(_ => sum < 1000).foreach(i => sum+=i)
(warning--this depends on details of how the takeWhile test and the foreach are interleaved during evaluation, and probably shouldn't be used in practice!).
(1b) Use tail recursion instead of a for loop, taking advantage of how easy it is to write a new method in Scala:
var sum = 0
def addTo(i: Int, max: Int) {
sum += i; if (sum < max) addTo(i+1,max)
}
addTo(0,1000)
(1c) Fall back to using a while loop
var sum = 0
var i = 0
while (i <= 1000 && sum <= 1000) { sum += 1; i += 1 }
(2) Throw an exception.
object AllDone extends Exception { }
var sum = 0
try {
for (i <- 0 to 1000) { sum += i; if (sum>=1000) throw AllDone }
} catch {
case AllDone =>
}
(2a) In Scala 2.8+ this is already pre-packaged in scala.util.control.Breaks using syntax that looks a lot like your familiar old break from C/Java:
import scala.util.control.Breaks._
var sum = 0
breakable { for (i <- 0 to 1000) {
sum += i
if (sum >= 1000) break
} }
(3) Put the code into a method and use return.
var sum = 0
def findSum { for (i <- 0 to 1000) { sum += i; if (sum>=1000) return } }
findSum
This is intentionally made not-too-easy for at least three reasons I can think of. First, in large code blocks, it's easy to overlook "continue" and "break" statements, or to think you're breaking out of more or less than you really are, or to need to break two loops which you can't do easily anyway--so the standard usage, while handy, has its problems, and thus you should try to structure your code a different way. Second, Scala has all sorts of nestings that you probably don't even notice, so if you could break out of things, you'd probably be surprised by where the code flow ended up (especially with closures). Third, most of Scala's "loops" aren't actually normal loops--they're method calls that have their own loop, or they are recursion which may or may not actually be a loop--and although they act looplike, it's hard to come up with a consistent way to know what "break" and the like should do. So, to be consistent, the wiser thing to do is not to have a "break" at all.
Note: There are functional equivalents of all of these where you return the value of sum rather than mutate it in place. These are more idiomatic Scala. However, the logic remains the same. (return becomes return x, etc.).
This has changed in Scala 2.8 which has a mechanism for using breaks. You can now do the following:
import scala.util.control.Breaks._
var largest = 0
// pass a function to the breakable method
breakable {
for (i<-999 to 1 by -1; j <- i to 1 by -1) {
val product = i * j
if (largest > product) {
break // BREAK!!
}
else if (product.toString.equals(product.toString.reverse)) {
largest = largest max product
}
}
}
It is never a good idea to break out of a for-loop. If you are using a for-loop it means that you know how many times you want to iterate. Use a while-loop with 2 conditions.
for example
var done = false
while (i <= length && !done) {
if (sum > 1000) {
done = true
}
}
To add Rex Kerr answer another way:
(1c) You can also use a guard in your loop:
var sum = 0
for (i <- 0 to 1000 ; if sum<1000) sum += i
Simply We can do in scala is
scala> import util.control.Breaks._
scala> object TestBreak {
def main(args : Array[String]) {
breakable {
for (i <- 1 to 10) {
println(i)
if (i == 5)
break;
} } } }
output :
scala> TestBreak.main(Array())
1
2
3
4
5
Since there is no break in Scala yet, you could try to solve this problem with using a return-statement. Therefore you need to put your inner loop into a function, otherwise the return would skip the whole loop.
Scala 2.8 however includes a way to break
http://www.scala-lang.org/api/rc/scala/util/control/Breaks.html
An approach that generates the values over a range as we iterate, up to a breaking condition, instead of generating first a whole range and then iterating over it, using Iterator, (inspired in #RexKerr use of Stream)
var sum = 0
for ( i <- Iterator.from(1).takeWhile( _ => sum < 1000) ) sum += i
// import following package
import scala.util.control._
// create a Breaks object as follows
val loop = new Breaks;
// Keep the loop inside breakable as follows
loop.breakable{
// Loop will go here
for(...){
....
// Break will go here
loop.break;
}
}
use Break module
http://www.tutorialspoint.com/scala/scala_break_statement.htm
Just use a while loop:
var (i, sum) = (0, 0)
while (sum < 1000) {
sum += i
i += 1
}
Here is a tail recursive version. Compared to the for-comprehensions it is a bit cryptic, admittedly, but I'd say its functional :)
def run(start:Int) = {
#tailrec
def tr(i:Int, largest:Int):Int = tr1(i, i, largest) match {
case x if i > 1 => tr(i-1, x)
case _ => largest
}
#tailrec
def tr1(i:Int,j:Int, largest:Int):Int = i*j match {
case x if x < largest || j < 2 => largest
case x if x.toString.equals(x.toString.reverse) => tr1(i, j-1, x)
case _ => tr1(i, j-1, largest)
}
tr(start, 0)
}
As you can see, the tr function is the counterpart of the outer for-comprehensions, and tr1 of the inner one. You're welcome if you know a way to optimize my version.
Close to your solution would be this:
var largest = 0
for (i <- 999 to 1 by -1;
j <- i to 1 by -1;
product = i * j;
if (largest <= product && product.toString.reverse.equals (product.toString.reverse.reverse)))
largest = product
println (largest)
The j-iteration is made without a new scope, and the product-generation as well as the condition are done in the for-statement (not a good expression - I don't find a better one). The condition is reversed which is pretty fast for that problem size - maybe you gain something with a break for larger loops.
String.reverse implicitly converts to RichString, which is why I do 2 extra reverses. :) A more mathematical approach might be more elegant.
I am new to Scala, but how about this to avoid throwing exceptions and repeating methods:
object awhile {
def apply(condition: () => Boolean, action: () => breakwhen): Unit = {
while (condition()) {
action() match {
case breakwhen(true) => return ;
case _ => { };
}
}
}
case class breakwhen(break:Boolean);
use it like this:
var i = 0
awhile(() => i < 20, () => {
i = i + 1
breakwhen(i == 5)
});
println(i)
if you don’t want to break:
awhile(() => i < 20, () => {
i = i + 1
breakwhen(false)
});
The third-party breakable package is one possible alternative
https://github.com/erikerlandson/breakable
Example code:
scala> import com.manyangled.breakable._
import com.manyangled.breakable._
scala> val bkb2 = for {
| (x, xLab) <- Stream.from(0).breakable // create breakable sequence with a method
| (y, yLab) <- breakable(Stream.from(0)) // create with a function
| if (x % 2 == 1) continue(xLab) // continue to next in outer "x" loop
| if (y % 2 == 0) continue(yLab) // continue to next in inner "y" loop
| if (x > 10) break(xLab) // break the outer "x" loop
| if (y > x) break(yLab) // break the inner "y" loop
| } yield (x, y)
bkb2: com.manyangled.breakable.Breakable[(Int, Int)] = com.manyangled.breakable.Breakable#34dc53d2
scala> bkb2.toVector
res0: Vector[(Int, Int)] = Vector((2,1), (4,1), (4,3), (6,1), (6,3), (6,5), (8,1), (8,3), (8,5), (8,7), (10,1), (10,3), (10,5), (10,7), (10,9))
import scala.util.control._
object demo_brk_963
{
def main(args: Array[String])
{
var a = 0;
var b = 0;
val numList1 = List(1,2,3,4,5,6,7,8,9,10);
val numList2 = List(11,12,13);
val outer = new Breaks; //object for break
val inner = new Breaks; //object for break
outer.breakable // Outer Block
{
for( a <- numList1)
{
println( "Value of a: " + a);
inner.breakable // Inner Block
{
for( b <- numList2)
{
println( "Value of b: " + b);
if( b == 12 )
{
println( "break-INNER;");
inner.break;
}
}
} // inner breakable
if( a == 6 )
{
println( "break-OUTER;");
outer.break;
}
}
} // outer breakable.
}
}
Basic method to break the loop, using Breaks class.
By declaring the loop as breakable.
Ironically the Scala break in scala.util.control.Breaks is an exception:
def break(): Nothing = { throw breakException }
The best advice is: DO NOT use break, continue and goto! IMO they are the same, bad practice and an evil source of all kind of problems (and hot discussions) and finally "considered be harmful". Code block structured, also in this example breaks are superfluous.
Our Edsger W. Dijkstra† wrote:
The quality of programmers is a decreasing function of the density of go to statements in the programs they produce.
I got a situation like the code below
for(id<-0 to 99) {
try {
var symbol = ctx.read("$.stocks[" + id + "].symbol").toString
var name = ctx.read("$.stocks[" + id + "].name").toString
stocklist(symbol) = name
}catch {
case ex: com.jayway.jsonpath.PathNotFoundException=>{break}
}
}
I am using a java lib and the mechanism is that ctx.read throw a Exception when it can find nothing.
I was trapped in the situation that :I have to break the loop when a Exception was thrown, but scala.util.control.Breaks.break using Exception to break the loop ,and it was in the catch block thus it was caught.
I got ugly way to solve this: do the loop for the first time and get the count of the real length.
and use it for the second loop.
take out break from Scala is not that good,when you are using some java libs.
Clever use of find method for collection will do the trick for you.
var largest = 0
lazy val ij =
for (i <- 999 to 1 by -1; j <- i to 1 by -1) yield (i, j)
val largest_ij = ij.find { case(i,j) =>
val product = i * j
if (product.toString == product.toString.reverse)
largest = largest max product
largest > product
}
println(largest_ij.get)
println(largest)
Below is code to break a loop in a simple way
import scala.util.control.Breaks.break
object RecurringCharacter {
def main(args: Array[String]) {
val str = "nileshshinde";
for (i <- 0 to str.length() - 1) {
for (j <- i + 1 to str.length() - 1) {
if (str(i) == str(j)) {
println("First Repeted Character " + str(i))
break() //break method will exit the loop with an Exception "Exception in thread "main" scala.util.control.BreakControl"
}
}
}
}
}
I don't know how much Scala style has changed in the past 9 years, but I found it interesting that most of the existing answers use vars, or hard to read recursion. The key to exiting early is to use a lazy collection to generate your possible candidates, then check for the condition separately. To generate the products:
val products = for {
i <- (999 to 1 by -1).view
j <- (i to 1 by -1).view
} yield (i*j)
Then to find the first palindrome from that view without generating every combination:
val palindromes = products filter {p => p.toString == p.toString.reverse}
palindromes.head
To find the largest palindrome (although the laziness doesn't buy you much because you have to check the entire list anyway):
palindromes.max
Your original code is actually checking for the first palindrome that is larger than a subsequent product, which is the same as checking for the first palindrome except in a weird boundary condition which I don't think you intended. The products are not strictly monotonically decreasing. For example, 998*998 is greater than 999*997, but appears much later in the loops.
Anyway, the advantage of the separated lazy generation and condition check is you write it pretty much like it is using the entire list, but it only generates as much as you need. You sort of get the best of both worlds.