The count is 4 at the end of the code below. I expected 0. Why is it 4? How can I get 0?
var count = 0;
"hello".forEach {
if(it == 'h')
{
println("Exiting the forEach loop. Count is $count");
return#forEach;
}
count++;
}
println("count is $count");
Output:
Exiting the forEach loop. Count is 0
count is 4
return#forEach does not exit forEach() itself, but the lambda passed to it ("body" of forEach()). Note that this lambda is executed several times - once per each item. By returning from it you actually skip only a single item, so this is similar to continue, not to break.
To workaround this you can create a label in the outer scope and return to it:
var count = 0;
run loop# {
"hello".forEach {
if(it == 'h')
{
println("Exiting the forEach loop. Count is $count");
return#loop;
}
count++;
}
}
This is described here: https://kotlinlang.org/docs/returns.html#return-at-labels
Note that the use of local returns in previous three examples is similar to the use of continue in regular loops. There is no direct equivalent for break, but it can be simulated by adding another nesting lambda and non-locally returning from it
It is 4 because the forEach call the lambda passed to it for each character in the string, so the return#forEach in your code return for the first element. You can use a for loop and use break to obtain 0.
return#forEach returns from the lambda function. But the forEach function is a higher-order function that calls the lambda repeatedly for each item in the iterator. So when you return from the lambda, you are only returning for that single item in the iterator. It is analogous to using continue in a traditional for loop.
If you want to exit iteration in a higher-order function completely, you have to use labels. And as I type this, I see another answer already shows how to do that, but I'll leave this in case the different explanation helps.
If your objective is to count the number of characters before 'h', you could do something like this:
val numCharsBeforeH = "hello".takeWhile { it != 'h' }.length
From your comment to Tenfour04's answer:
This is not very convenient. Why didn't the makers of Kotlin create a "break" equivalent?
Here is a quote of the "Loops" section of the Coding conventions:
Prefer using higher-order functions (filter, map etc.) to loops.
Exception: forEach (prefer using a regular for loop instead, unless
the receiver of forEach is nullable or forEach is used as part of a
longer call chain).
When making a choice between a complex expression using multiple
higher-order functions and a loop, understand the cost of the
operations being performed in each case and keep performance
considerations in mind.
Indeed, using a regular for loop with break does what you expect:
var count = 0;
for (char in "hello") {
if (char == 'h') {
println("Breaking the loop. Count is $count")
break
}
count++
}
println("count is $count")
Output:
Breaking the loop. Count is 0
count is 0
Except for very simple operations, there are probably better ways to do what you need than using forEach.
I am trying to solve the following question on LeetCode; Write a function that takes an unsigned integer and returns the number of '1' bits it has. Constraints: The input must be a binary string of length 32.
I have written the following code for that which works fine for inputs 00000000000000000000000000001011 and 00000000000000000000000010000000 (provided internally by the website) but give output 0 for input 11111111111111111111111111111101 and in my local compiler for the last input it says "out of range"
class Solution {
// you need treat n as an unsigned value
fun hammingWeight(n:Int):Int {
var num = n
var setCountBit = 0
while (num > 0) {
setCountBit++
num= num and num-1
}
return setCountBit
}
}
To correctly convert binary string to Int and avoid "out of range error", you need to do the following (I believe LeetCode does the same under the hood):
fun binaryStringToInt(s: String): Int = s.toUInt(radix = 2).toInt()
"11111111111111111111111111111101" is equivalent to 4294967293. This is greater than Int.MAX_VALUE, so it will be represented as negative number after .toInt() convertion (-3 in this case).
Actually, this problem could be solved with one-liner in Kotlin 1.4:
fun hammingWeight(n: Int): Int = n.countOneBits()
But LeetCode uses Kotlin 1.3.10, so you need to adjust your solution to handle negative Ints as well.
Please change the type of your input variable from Int to a type like Double .At the moment The given value is bigger than the maximum value that a type Int number can store.
In the code below:
var verticesCount: Int // to read a vertices count for graph
// Reading until we get a valid vertices count.
while (!Assertions.checkEnoughVertices(
verticesCount = consoleReader.readInt(null, Localization.getLocStr("type_int_vertices_count"))))
// The case when we don't have enough vertices.
println(String.format(Localization.getLocStr("no_enough_vertices_in_graph"),
Assertions.CONFIG_MIN_VERTICES_COUNT))
val resultGraph = Graph(verticesCount)
we are getting next error on the last line:
Error:(31, 33) Kotlin: Variable 'verticesCount' must be initialized
Assertions.checkEnoughVertices accepts a safe type variable as an argument (verticesCount: Int), so it's impossible for verticesCount to be uninitialized or null here (and we're getting no corresponding errors on those lines).
What's going on on the last line when already initialized variable becomes uninitialized again?
The syntax you've used denotes a function call with named arguments, not the assignment of a local variable. So verticesCount = is just an explanation to the reader that the value which is being passed here to checkEnoughVertices corresponds to the parameter of that function named verticesCount. It has nothing to do with the local variable named verticesCount declared just above, so the compiler thinks you've still to initialize that variable.
In Kotlin, the assignment to a variable (a = b) is not an expression, so it cannot be used as a value in other expressions. You have to split the assignment and the while-loop condition to achieve what you want. I'd do this with an infinite loop + a condition inside:
var verticesCount: Int
while (true) {
verticesCount = consoleReader.readInt(...)
if (Assertions.checkEnoughVertices(verticesCount)) break
...
}
val resultGraph = Graph(verticesCount)
Well, technically it is possible to assign values to variables in the while condition - and anything else you might want to do there, too.
The magic comes from the also function:
Try this: (excuse the completely useless thing this is doing...)
var i = 10
var doubleI: Int
while ((i * 2).also { doubleI = it } > 0) {
i--
println(doubleI)
}
Any expression can be "extended" with "something to do" by calling also which takes the expression it is called upon as the it parameter and executes the given block. The value also returns is identical to its caller value.
Here's a very good article to explain this and much more: https://medium.com/#elye.project/mastering-kotlin-standard-functions-run-with-let-also-and-apply-9cd334b0ef84
I'm reading a book to learn Objective-C and this program is suppose to show key concepts in dealing with pointers, and I'm really lost.
Is there some kind of conversion happening in the function's arguments that turn p1, p2, &il, and &i2 to the value (*) of a pointer? Like p1 turns into *p1?
I thought a copy of the variable was passed into the function instead of the actual variable, so why was the value of the passed in variable changed after the function?
Also why am I getting a warning on the 3rd line that says: No previous prototype for function 'exchangeValues'?
Thank you!!
#import <Foundation/Foundation.h>
void exchangeValues (int *pint1, int *pint2) {
int temp;
temp = *pint1;
*pint1 = *pint2;
*pint2 = temp;
}
int main (int argc, char *argv[]) {
#autoreleasepool {
void exchangeValues (int *pint1, int *pint2);
int il = -5, i2 = 66, *p1 = &il, *p2 = &i2;
NSLog(#"il = %i, i2 = %i", il, i2);
exchangeValues(p1, p2);
NSLog(#"il = %i, i2 = %i", il, i2);
exchangeValues(&il, &i2);
NSLog(#"il = %i, i2 = %i", il, i2);
}
return 0;
}
Output:
2012-08-02 11:13:38.569 Test[381:707] il = -5, i2 = 66
2012-08-02 11:13:38.571 Test[381:707] il = 66, i2 = -5
2012-08-02 11:13:38.572 Test[381:707] il = -5, i2 = 66
I would say that's a complex example if you are being taught about pointers!
Is there some kind of conversion happening in the function's arguments
that turn p1, p2, &il, and &i2 to the value (*) of a pointer? Like p1
turns into *p1?
p1 and p2 are declared as int * (pointer to int) and are initialised with the address of i1 and i2 (using the & operator).
I thought a copy of the variable was passed into the function instead
of the actual variable, so why was the value of the passed in variable
changed after the function?
A copy of the variable is passed to the function, however in this case the variable of type int * (pointer to int). The reason the value is changing is because the exchangeValues() function is dereferencing those pointers and swapping the values. This is the only way (in C/Objective-C) a function can modify a variable outside of its own scope, other than the variable being assigned as the return value from a function.
Also why am I getting a warning on the 3rd line that says: No previous
prototype for function 'exchangeValues'?
You seem to have typed it in wrong; remove the line below #autoreleasepool:
#autoreleasepool {
void exchangeValues (int *pint1, int *pint2); <-- delete this line
If you pass a pointer into the function, it indeed passes a copy of that pointer- but it still refers to the same address in memory. So de-referencing that pointer will still point to a variable that's outside of the function scope.
I thought a copy of the variable was passed into the function instead of the actual variable, so why was the value of the passed in variable changed after the function?
A copy of the pointer is passed to the function here. So what the function has points to the memory locations the variables l1 and l2 are stored at. So
void exchangeValues (int *pint1, int *pint2) {
int temp;
temp = *pint1; // store the value that pint1 points to in temp
*pint1 = *pint2; // store the value pint2 points to where pint1 points to
*pint2 = temp; // store the value saved in temp where pint2 points to
}
its a little confusing how the variables have been declared and initialised all in a row like that but basically you have:
i1 is an int set to -5
p1 is a pointer to an int set to the address of i1
same goes for i2 and p2
No conversion is taking place. You're effectively 'swapping' the values that those pointers point to in the function.
Pointers are confusing things but stick with it and it will become clear with enough parctice and example code like this...
I'm studying dynamic/static scope with deep/shallow binding and running code manually to see how these different scopes/bindings actually work. I read the theory and googled some example exercises and the ones I found are very simple (like this one which was very helpful with dynamic scoping) But I'm having trouble understanding how static scope works.
Here I post an exercise I did to check if I got the right solution:
considering the following program written in pseudocode:
int u = 42;
int v = 69;
int w = 17;
proc add( z:int )
u := v + u + z
proc bar( fun:proc )
int u := w;
fun(v)
proc foo( x:int, w:int )
int v := x;
bar(add)
main
foo(u,13)
print(u)
end;
What is printed to screen
a) using static scope? answer=180
b) using dynamic scope and deep binding? answer=69 (sum for u = 126 but it's foo's local v, right?)
c) using dynamic scope and shallow binding? answer=69 (sum for u = 101 but it's foo's local v, right?)
PS: I'm trying to practice doing some exercises like this if you know where I can find these types of problems (preferable with solutions) please give the link, thanks!
Your answer for lexical (static) scope is correct. Your answers for dynamic scope are wrong, but if I'm reading your explanations right, it's because you got confused between u and v, rather than because of any real misunderstanding about how deep and shallow binding work. (I'm assuming that your u/v confusion was just accidental, and not due to a strange confusion about values vs. references in the call to foo.)
a) using static scope? answer=180
Correct.
b) using dynamic scope and deep binding? answer=69 (sum for u = 126 but it's foo's local v, right?)
Your parenthetical explanation is right, but your answer is wrong: u is indeed set to 126, and foo indeed localizes v, but since main prints u, not v, the answer is 126.
c) using dynamic scope and shallow binding? answer=69 (sum for u = 101 but it's foo's local v, right?)
The sum for u is actually 97 (42+13+42), but since bar localizes u, the answer is 42. (Your parenthetical explanation is wrong for this one — you seem to have used the global variable w, which is 17, in interpreting the statement int u := w in the definition of bar; but that statement actually refers to foo's local variable w, its second parameter, which is 13. But that doesn't actually affect the answer. Your answer is wrong for this one only because main prints u, not v.)
For lexical scope, it's pretty easy to check your answers by translating the pseudo-code into a language with lexical scope. Likewise dynamic scope with shallow binding. (In fact, if you use Perl, you can test both ways almost at once, since it supports both; just use my for lexical scope, then do a find-and-replace to change it to local for dynamic scope. But even if you use, say, JavaScript for lexical scope and Bash for dynamic scope, it should be quick to test both.)
Dynamic scope with deep binding is much trickier, since few widely-deployed languages support it. If you use Perl, you can implement it manually by using a hash (an associative array) that maps from variable-names to scalar-refs, and passing this hash from function to function. Everywhere that the pseudocode declares a local variable, you save the existing scalar-reference in a Perl lexical variable, then put the new mapping in the hash; and at the end of the function, you restore the original scalar-reference. To support the binding, you create a wrapper function that creates a copy of the hash, and passes that to its wrapped function. Here is a dynamically-scoped, deeply-binding implementation of your program in Perl, using that approach:
#!/usr/bin/perl -w
use warnings;
use strict;
# Create a new scalar, initialize it to the specified value,
# and return a reference to it:
sub new_scalar($)
{ return \(shift); }
# Bind the specified procedure to the specified environment:
sub bind_proc(\%$)
{
my $V = { %{+shift} };
my $f = shift;
return sub { $f->($V, #_); };
}
my $V = {};
$V->{u} = new_scalar 42; # int u := 42
$V->{v} = new_scalar 69; # int v := 69
$V->{w} = new_scalar 17; # int w := 17
sub add(\%$)
{
my $V = shift;
my $z = $V->{z}; # save existing z
$V->{z} = new_scalar shift; # create & initialize new z
${$V->{u}} = ${$V->{v}} + ${$V->{u}} + ${$V->{z}};
$V->{z} = $z; # restore old z
}
sub bar(\%$)
{
my $V = shift;
my $fun = shift;
my $u = $V->{u}; # save existing u
$V->{u} = new_scalar ${$V->{w}}; # create & initialize new u
$fun->(${$V->{v}});
$V->{u} = $u; # restore old u
}
sub foo(\%$$)
{
my $V = shift;
my $x = $V->{x}; # save existing x
$V->{x} = new_scalar shift; # create & initialize new x
my $w = $V->{w}; # save existing w
$V->{w} = new_scalar shift; # create & initialize new w
my $v = $V->{v}; # save existing v
$V->{v} = new_scalar ${$V->{x}}; # create & initialize new v
bar %$V, bind_proc %$V, \&add;
$V->{v} = $v; # restore old v
$V->{w} = $w; # restore old w
$V->{x} = $x; # restore old x
}
foo %$V, ${$V->{u}}, 13;
print "${$V->{u}}\n";
__END__
and indeed it prints 126. It's obviously messy and error-prone, but it also really helps you understand what's going on, so for educational purposes I think it's worth it!
Simple and deep binding are Lisp interpreter viewpoints of the pseudocode. Scoping is just pointer arithmetic. Dynamic scope and static scope are the same if there are no free variables.
Static scope relies on a pointer to memory. Empty environments hold no symbol to value associations; denoted by word "End." Each time the interpreter reads an assignment, it makes space for association between a symbol and value.
The environment pointer is updated to point to the last association constructed.
env = End
env = [u,42] -> End
env = [v,69] -> [u,42] -> End
env = [w,17] -> [v,69] -> [u,42] -> End
Let me record this environment memory location as AAA. In my Lisp interpreter, when meeting a procedure, we take the environment pointer and put it our pocket.
env = [add,[closure,(lambda(z)(setq u (+ v u z)),*AAA*]]->[w,17]->[v,69]->[u,42]->End.
That's pretty much all there is until the procedure add is called. Interestingly, if add is never called, you just cost yourself a pointer.
Suppose the program calls add(8). OK, let's roll. The environment AAA is made current. Environment is ->[w,17]->[v,69]->[u,42]->End.
Procedure parameters of add are added to the front of the environment. The environment becomes [z,8]->[w,17]->[v,69]->[u,42]->End.
Now the procedure body of add is executed. Free variable v will have value 69. Free variable u will have value 42. z will have the value 8.
u := v + u + z
u will be assigned the value of 69 + 42 + 8 becomeing 119.
The environment will reflect this: [z,8]->[w,17]->[v,69]->[u,119]->End.
Assume procedure add has completed its task. Now the environment gets restored to its previous value.
env = [add,[closure,(lambda(z)(setq u (+ v u z)),*AAA*]]->[w,17]->[v,69]->[u,119]->End.
Notice how the procedure add has had a side effect of changing the value of free variable u. Awesome!
Regarding dynamic scoping: it just ensures closure leaves out dynamic symbols, thereby avoiding being captured and becoming dynamic.
Then put assignment to dynamic at top of code. If dynamic is same as parameter name, it gets masked by parameter value passed in.
Suppose I had a dynamic variable called z. When I called add(8), z would have been set to 8 regardless of what I wanted. That's probably why dynamic variables have longer names.
Rumour has it that dynamic variables are useful for things like backtracking, using let Lisp constructs.
Static binding, also known as lexical scope, refers to the scoping mechanism found in most modern languages.
In "lexical scope", the final value for u is neither 180 or 119, which are wrong answers.
The correct answer is u=101.
Please see standard Perl code below to understand why.
use strict;
use warnings;
my $u = 42;
my $v = 69;
my $w = 17;
sub add {
my $z = shift;
$u = $v + $u + $z;
}
sub bar {
my $fun = shift;
$u = $w;
$fun->($v);
}
sub foo {
my ($x, $w) = #_;
$v = $x;
bar( \&add );
}
foo($u,13);
print "u: $u\n";
Regarding shallow binding versus deep binding, both mechanisms date from the former LISP era.
Both mechanisms are meant to achieve dynamic binding (versus lexical scope binding) and therefore they produce identical results !
The differences between shallow binding and deep binding do not reside in semantics, which are identical, but in the implementation of dynamic binding.
With deep binding, variable bindings are set within a stack as "varname => varvalue" pairs.
The value of a given variable is retrieved from traversing the stack from top to bottom until a binding for the given variable is found.
Updating the variable consists in finding the binding in the stack and updating the associated value.
On entering a subroutine, a new binding for each actual parameter is pushed onto the stack, potentially hiding an older binding which is therefore no longer accessible wrt the retrieving mechanism described above (that stops at the 1st retrieved binding).
On leaving the subroutine, bindings for these parameters are simply popped from the binding stack, thus re-enabling access to the former bindings.
Please see the the code below for a Perl implementation of deep-binding dynamic scope.
use strict;
use warnings;
use utf8;
##
# Dynamic-scope deep-binding implementation
my #stack = ();
sub bindv {
my ($varname, $varval);
unshift #stack, [ $varname => $varval ]
while ($varname, $varval) = splice #_, 0, 2;
return $varval;
}
sub unbindv {
my $n = shift || 1;
shift #stack while $n-- > 0;
}
sub getv {
my $varname = shift;
for (my $i=0; $i < #stack; $i++) {
return $stack[$i][1]
if $varname eq $stack[$i][0];
}
return undef;
}
sub setv {
my ($varname, $varval) = #_;
for (my $i=0; $i < #stack; $i++) {
return $stack[$i][1] = $varval
if $varname eq $stack[$i][0];
}
return bindv($varname, $varval);
}
##
# EXERCICE
bindv( u => 42,
v => 69,
w => 17,
);
sub add {
bindv(z => shift);
setv(u => getv('v')
+ getv('u')
+ getv('z')
);
unbindv();
}
sub bar {
bindv(fun => shift);
setv(u => getv('w'));
getv('fun')->(getv('v'));
unbindv();
}
sub foo {
bindv(x => shift,
w => shift,
);
setv(v => getv('x'));
bar( \&add );
unbindv(2);
}
foo( getv('u'), 13);
print "u: ", getv('u'), "\n";
The result is u=97
Nevertheless, this constant traversal of the binding stack is costly : 0(n) complexity !
Shallow binding brings a wonderful O(1) enhanced performance over the previous implementation !
Shallow binding is improving the former mechanism by assigning each variable its own "cell", storing the value of the variable within the cell.
The value of a given variable is simply retrieved from the variable's
cell (using a hash table on variable names, we achieve a
0(1) complexity for accessing variable's values!)
Updating the variable's value is simply storing the value into the
variable's cell.
Creating a new binding (entering subs) works by pushing the old value
of the variable (a previous binding) onto the stack, and storing the
new local value in the value cell.
Eliminating a binding (leaving subs) works by popping the old value
off the stack into the variable's value cell.
Please see the the code below for a trivial Perl implementation of shallow-binding dynamic scope.
use strict;
use warnings;
our $u = 42;
our $v = 69;
our $w = 17;
our $z;
our $fun;
our $x;
sub add {
local $z = shift;
$u = $v + $u + $z;
}
sub bar {
local $fun = shift;
$u = $w;
$fun->($v);
}
sub foo {
local $x = shift;
local $w = shift;
$v = $x;
bar( \&add );
}
foo($u,13);
print "u: $u\n";
As you shall see, the result is still u=97
As a conclusion, remember two things :
shallow binding produces the same results as deep binding, but runs faster, since there is never a need to search for a binding.
The problem is not shallow binding versus deep binding versus
static binding BUT lexical scope versus dynamic scope (implemented either with deep or shallow binding).