I'm trying to create a loop to read, for example, 4200 users from 1000 to 1000 but I can't get it to cut when it reaches the end. I tried it with if, for and I couldn't do it.
I have programmed in JAVA but with Groovy I see that the structure is different.
urlUsers = urlUsers.concat("/1/1000");
List<UserConnectorObject> usersList = null;
while({
gesdenResponse = GesdenUtils.sendHttpRequest(urlUsers, "LOOKUP", null,
request.getMetaData()?.getLogin(), request.getMetaData()?.getPassword());
log.info("Users data in JSON: "+gesdenResponse.getOutput())
usersList = GesdenUtils.fromJSON(gesdenResponse.getOutput(), GesdenConstants.USER_IDENTITY_KEY);
usersList.size() == 10;
log.info("List size in JSON "+usersList.size());
}()) continue
Groovy has lots of loop structures, but it is crucial to separate the regular ones (lang built-ins) and the api functions which take closure as an argument
take closure - no plain way to escape
If you want to iterate from A to B users, you can use, for instance,
(10..20).each { userNo -> // Here you will have all 10 iterations
if ( userNo == 5) {
return
}
}
If something outrageous happens in the loop body and you cannot use return to escape, as loop boddy is a closure (separate function) and this resurn just exits this closure. Next iteration will happen just after.
use regular lang built-in loop structures - make use of break/continue
for (int userNo in 1..10) { // Here you will have only 5 iterations
if (userNo == 5) {
break
}
}
It looks like your closure always return falsy because there is no explicit return, and the last statement evaluated is the call to log.info(String) which returns void.
Use an explicit return or move/delete the log statement.
Related
I have a function that takes a vector as a parameter, scan this vector and generates a random word. It's expected from me that the generated words' letters are different from each other. So, I want to check it with a simple if-else condition inside the same function. If all letters are different, function returns this word. If not, I need to use the same function which I am already inside while using conditions. But first parameter that I used in the main function doesn't work when I attempt to use it for the second time. Here the generateaRandomWord(vector a) function:
vector<string> currentVector;
string generateaRandomWord(vector<string> a) {
currentVector = a;
string randomWord;
int randomNumber = rand() % currentVector.size();
randomWord = currentVector.at(randomNumber);
if (hasUniqueChars(randomWord)) {
return randomWord;
}
else {
generateaRandomWord(currentVector);
}
}
I thought that it is a good idea to keep a vector (currentVector) outside of the function. So, for the first time I use the function this vector will be defined and I will be able to use it if using recursion is necessary. But that didn't work either.
The main problem you have is that your recursive case doesn't return anything -- it throws away the returned value from the recursive call, then falls off the end of the function (returning garbage -- undefined behvaior). You need to actually return the value returned by the recursive call:
return generateaRandomWord(currentVector);
I've been learning lua and can't seem to make a simple implementation of this binary tree work...
function createTree(tree, max)
if max > 0 then
tree = {data = max, left = {}, right = {}}
createTree(tree.left, max - 1)
createTree(tree.right, max - 1)
end
end
function printTree(tree)
if tree then
print(tree.data)
printTree(tree.left)
printTree(tree.right)
end
end
tree = {}
createTree(tree, 3)
printTree(tree)
the program just returns nil after execution. I've searched around the web to understand how argument passing works in lua (if it is by reference or by value) and found out that some types are passed by reference (like tables and functions) while others by value. Still, I made the global variable "tree" a table before passing it to the "createTree" function, and I even initialized "left" and "right" to be empty tables inside of "createTree" for the same purpose. What am I doing wrong?
It is probably necessary to initialize not by a new table, but only to set its values.
function createTree(tree, max)
if max > 0 then
tree.data = max
tree.left = {}
tree.right = {}
createTree(tree.left, max - 1)
createTree(tree.right, max - 1)
end
end
in Lua, arguments are passed by value. Assigning to an argument does not change the original variable.
Try this:
function createTree(max)
if max == 0 then
return nil
else
return {data = max, left = createTree(max-1), right = createTree(max-1)}
end
end
It is safe to think that for the most of the cases lua passes arguments by value. But for any object other than a number (numbers aren't objects actually), the "value" is actually a pointer to the said object.
When you do something like a={1,2,3} or b="asda" the values on the right are allocated somewhere dynamically, and a and b only get addresses of those. Thus, when you pass a to the function fun(a), the pointer is copied to a new variable inside function, but the a itself is unaffected:
function fun(p)
--p stores address of the same object, but `p` is not `a`
p[1]=3--by using the address you can
p[4]=1--alter the contents of the object
p[2]=nil--this will be seen outside
q={}
p={}--here you assign address of another object to the pointer
p=q--(here too)
end
Functions are also represented by pointers to them, you can use debug library to tinker with function object (change upvalues for example), this may affect how function executes, but, once again, you can not change where external references are pointing.
Strings are immutable objects, you can pass them around, there is a library that does stuff to them, but all the functions in that library return new string. So once, again external variable b from b="asda" would not be affected if you tried to do something with "asda" string inside the function.
I was wondering why this is valid go code:
func FindUserInfo(id string) (Info, bool) {
it, present := all[id]
return it, present
}
but this isn't
func FindUserInfo(id string) (Info, bool) {
return all[id]
}
is there a way to avoid the temporary variables?
To elaborate on my comment, the Effective Go mentions that the multi-value assignment from accessing a map key is called the "comma ok" pattern.
Sometimes you need to distinguish a missing entry from a zero value. Is there an entry for "UTC" or is that the empty string because it's not in the map at all? You can discriminate with a form of multiple assignment.
var seconds int
var ok bool
seconds, ok = timeZone[tz]
For obvious reasons this is called the “comma ok” idiom. In this example, if tz is present, seconds will be set appropriately and ok will be true; if not, seconds will be set to zero and ok will be false.
Playground demonstrating this
We can see that this differs from calling a regular function where the compiler would tell you that something is wrong:
package main
import "fmt"
func multiValueReturn() (int, int) {
return 0, 0
}
func main() {
fmt.Println(multiValueReturn)
asgn1, _ := multiValueReturn()
asgn2 := multiValueReturn()
}
On the playground this will output
# command-line-arguments
/tmp/sandbox592492597/main.go:14: multiple-value multiValueReturn() in single-value context
This gives us a hint that it may be something the compiler is doing. Searching the source code for "commaOk" gives us a few places to look, including types.unpack
At the time of writing this it this the method's godoc reads:
// unpack takes a getter get and a number of operands n. If n == 1, unpack
// calls the incoming getter for the first operand. If that operand is
// invalid, unpack returns (nil, 0, false). Otherwise, if that operand is a
// function call, or a comma-ok expression and allowCommaOk is set, the result
// is a new getter and operand count providing access to the function results,
// or comma-ok values, respectively. The third result value reports if it
// is indeed the comma-ok case. In all other cases, the incoming getter and
// operand count are returned unchanged, and the third result value is false.
//
// In other words, if there's exactly one operand that - after type-checking
// by calling get - stands for multiple operands, the resulting getter provides
// access to those operands instead.
//
// If the returned getter is called at most once for a given operand index i
// (including i == 0), that operand is guaranteed to cause only one call of
// the incoming getter with that i.
//
The key bits of this being that this method appears to determine whether or not something is actually a "comma ok" case.
Digging into that method tells us that it will check to see if the mode of the operands is indexing a map or if the mode is set to commaok (where this is defined does give us many hints on when it's used, but searching the source for assignments to commaok we can see it's used when getting a value from a channel and type assertions). Remember the bolded bit for later!
if x0.mode == mapindex || x0.mode == commaok {
// comma-ok value
if allowCommaOk {
a := [2]Type{x0.typ, Typ[UntypedBool]}
return func(x *operand, i int) {
x.mode = value
x.expr = x0.expr
x.typ = a[i]
}, 2, true
}
x0.mode = value
}
allowCommaOk is a parameter to the function. Checking out where unpack is called in that file we can see that all callers pass false as an argument. Searching the rest of the repository leads us to assignments.go in the Checker.initVars() method.
l := len(lhs)
get, r, commaOk := unpack(func(x *operand, i int) { check.expr(x, rhs[i]) }, len(rhs), l == 2 && !returnPos.IsValid())
Since it seems that we can only use the "comma ok" pattern to get two return values when doing a multi-value assignment this seems like the right place to look! In the above code the length of the left hand side is checked, and when unpack is called the allowCommaOk parameter is the result of l == 2 && !returnPos.IsValid(). The !returnPos.IsValid() is somewhat confusing here as that would mean that the position has no file or line information associated with it, but we'll just ignore that.
Further down in that method we've got:
var x operand
if commaOk {
var a [2]Type
for i := range a {
get(&x, i)
a[i] = check.initVar(lhs[i], &x, returnPos.IsValid())
}
check.recordCommaOkTypes(rhs[0], a)
return
}
So what does all of this tell us?
Since the unpack method takes an allowCommaOk parameter that's hardcoded to false everywhere except in assignment.go's Checker.initVars() method, we can probably assume that you will only ever get two values when doing an assignment and have two variables on the left-hand side.
The unpack method will determine whether or not you actually do get an ok value in return by checking if you are indexing a slice, grabbing a value from a channel, or doing a type assertion
Since you can only get the ok value when doing an assignment it looks like in your specific case you will always need to use variables
You may save a couple of key strokes by using named returns:
func FindUserInfo(id string) (i Info, ok bool) {
i, ok = all[id]
return
}
But apart from that, I don't think what you want is possible.
Simply put: the reason why your second example isn't valid Go code is because the language specification says so. ;)
Indexing a map only yields a secondary value in an assignment to two variables. Return statement is not an assignment.
An index expression on a map a of type map[K]V used in an assignment or initialization of the special form
v, ok = a[x]
v, ok := a[x]
var v, ok = a[x]
yields an additional untyped boolean value. The value of ok is true if the key x is present in the map, and false otherwise.
Furthermore, indexing a map is not a "single call to a multi-valued function", which is one of the three ways to return values from a function (the second one, the other two not being relevant here):
There are three ways to return values from a function with a result type:
The return value or values may be explicitly listed in the "return" statement. Each expression must be single-valued and assignable to the corresponding element of the function's result type.
The expression list in the "return" statement may be a single call to a multi-valued function. The effect is as if each value returned from that function were assigned to a temporary variable with the type of the respective value, followed by a "return" statement listing these variables, at which point the rules of the previous case apply.
The expression list may be empty if the function's result type specifies names for its result parameters. The result parameters act as ordinary local variables and the function may assign values to them as necessary. The "return" statement returns the values of these variables.
As for your actual question: the only way to avoid temporary variables would be using non-temporary variables, but usually that would be quite unwise - and probably not much of an optimization even when safe.
So, why doesn't the language specification allow this kind of special use of map indexing (or type assertion or channel receive, both of which can also utilize the "comma ok" idiom) in return statements? That's a good question. My guess: to keep the language specification simple.
I'm no Go expert but I believe you are getting compile time error when you are trying to return the array i.e. return all[id]. The reason could be because the functions return type is specially mentioned as (Info, bool) and when you are doing return all[id] it can't map the return type of all[id] to (Info, bool).
However the solution mentioned above, the variables being returned i and ok are the same that are mentioned in the return type of the function (i Info, ok bool) and hence the compiler knows what it's returning as opposed to just doing (i Info, ok bool).
By default, maps in golang return a single value when accessing a key
https://blog.golang.org/go-maps-in-action
Hence, return all[id] won't compile for a function that expects 2 return values.
Here is coding from Couchbase Document and I dont understand it
function(key, values, rereduce) {
var result = {total: 0, count: 0};
for(i=0; i < values.length; i++) {
if(rereduce) {
result.total = result.total + values[i].total;
result.count = result.count + values[i].count;
} else {
result.total = sum(values);
result.count = values.length;
}
}
return(result);
}
rereduce means the current function call has already done the reduce or not. right?
the first argument of the reduce function, key, when will it be used? I saw a numbers of examples, key seems to be unused
When does rereduce return true and the array size is more than 1?
Again, When does rereduce return is false and the array size is more than 1?
Rereduce means that the reduce function is called before and now it is called again with params that were returnd as a result in first reduce call. So if we devide it into two functions it will look like:
function reduce(k,v){
// ... doing something with map results
// instead of returning result we must call rereduce function)
rereduce(null, result)
}
function rereduce(k,v){
// do something with first reduce result
}
In most cases rereduce will happen when you have 2 or more servers in cluster or you have a lot of items in your database and the calculation is done on multiple "nodes" of the B*Tree. Example with 2 servers will be easier to understand:
Let's imagine that your map function returned pairs: [key1-1, key2-2, key6-6] from 1st server and [key5-5,key7-7] from 2nd. You'll get 2 reduce function calls with:
reduce([key1,key2,key6],[1,2,6],false) and reduce([key5,key7],[5,7],false). Then if we just return values (do nothing in reduce, just return values), the reduce function will be called with such params: reduce(null, [[1,2,6],[5,7]], true). Here values will be an array of results that came from first reduce calls.
On rereduce key will be null. Values will be an array of values as returned by a previous reduce() function.
Array size depends only on your data. It not depends on rereduce variable. Same answer for 4th question.
You can just try to run examples from Views basics and Views with reduce. I.e. you can modify reduce function to see what it returns on each step:
function reduce(k,v,r){
if (!r){
// let reduce function return only one value:
return 1;
} else {
// and lets see what values have came in "rereduce"
return v;
}
}
I am also confused by the example from the official couchbase website as well, and below is what i thought.
confusion: the reduce method signature
1) its written as
function (keys, values, rereduce)
2) its written as function(key, values, rereduce)
What exactly is the first param, key or keys
For all my understand from my previous exp on the map/reduce, the key the key that emit from the map function and there is a hidden shuffle method that will aggregate the value into a value list for the same key.
So the key param can be an array under the circumstances that you emit an array as key (which you can use group by level control the level of aggregation)
So i am not agree with the example that given by #m03geek, it should not be a list of different keys, correct me if i am wrong.
My assumption:
Both reduce and rereduce work on the SAME key only.
eg:
reduce is like:
1)reduce(keyA, [1,2,3]) this is precalculated, and stored in Btree structure
2) rereduce(keyA, [6, reduce(keyA, [4,5,6])]), 6 is the sum of [1,2,3] from the first reduce method, then we add a new doc into couchbase, which will trigger the reduce method again, instead of calculating the whole thing again as the original map/reduce will do, couchbase get the precalculated data out from the btree which is 6, and run reduce from the key-value pairs from the map method (which is triggered by adding a new doc), then run re-reduce on the precalculated value + new value.
I was able to store functions into a table. But now I have no idea of how to invoke them. The final table will have about 100 calls, so if possible, I'd like to invoke them as if in a foreach loop. Thanks!
Here is how the table was defined:
game_level_hints = game_level_hints or {}
game_level_hints.levels = {}
game_level_hints.levels["level0"] = function()
return
{
[on_scene("scene0")] =
{
talk("hint0"),
talk("hint1"),
talk("hint2")
},
[on_scene("scene1")] =
{
talk("hint0"),
talk("hint1"),
talk("hint2")
}
}
end
Aaand the function definitions:
function on_scene(sceneId)
-- some code
return sceneId
end
function talk(areaId)
-- some code
return areaId
end
EDIT:
I modified the functions so they'll have a little more context. Basically, they return strings now. And what I was hoping to happen is that at then end of invoking the functions, I'll have a table (ideally the levels table) containing all these strings.
Short answer: to call a function (reference) stored in an array, you just add (parameters), as you'd normally do:
local function func(a,b,c) return a,b,c end
local a = {myfunc = func}
print(a.myfunc(3,4,5)) -- prints 3,4,5
In fact, you can simplify this to
local a = {myfunc = function(a,b,c) return a,b,c end}
print(a.myfunc(3,4,5)) -- prints 3,4,5
Long answer: You don't describe what your expected results are, but what you wrote is likely not to do what you expect it to do. Take this fragment:
game_level_hints.levels["level0"] = function()
return
{
[on_scene("scene0")] =
{
talk("hint0"),
}
}
end
[This paragraph no longer applies after the question has been updated] You reference on_scene and talk functions, but you don't "store" those functions in the table (since you explicitly referenced them in your question, I presume the question is about these functions). You actually call these functions and store the values they return (they both return nil), so when this fragment is executed, you get "table index is nil" error as you are trying to store nil using nil as the index.
If you want to call the function you stored in game_level_hints.levels["level0"], you just do game_level_hints.levels["level0"]()
Using what you guys answered and commented, I was able to come up with the following code as a solution:
asd = game_level_hints.levels["level0"]()
Now, asd contains the area strings I need. Although ideally, I intended to be able to access the data like:
asd[1][1]
accessing it like:
asd["scene0"][1]
to retrieve the area data would suffice. I'll just have to work around the keys.
Thanks, guys.
It's not really clear what you're trying to do. Inside your anonymous function, you're returning a table that uses on_scene's return value as keys. But your on_scene doesn't return anything. Same thing for talk.
I'm going to assume that you wanted on_scene and talk to get called when invoking each levels in your game_level_hints table.
If so, this is how you can do it:
local maxlevel = 99
for i = 0, maxlevel do
game_level_hints.levels["level" .. i] = function()
on_scene("scene" .. i)
talk("hint" .. i)
end
end
-- ...
for levelname, levelfunc in pairs(game_level_hints.levels) do
levelfunc()
end