is there anyway to declare a parameter as modifiable without

Question

is there anyway to declare a parameter as modifiable without needing to do ``` fun a b Int var c = b c = 5 ```

diesieben07 · Answer

No parameters are always `val`

neil.armstrong · Answer

Can do shadowing though not really recommended ``` fun a b Int var b = b b = 5 ```

diesieben07 · Answer

In general if you want to have `var` semantics for a parameter that indicates a code smell

Smallville7123 · Answer

so modifying a parameter passed value is undefined behaviour and it should not be done

diesieben07 · Answer

It s not undefined behavior It is considered a code smell so it is not allowed in Kotlin What is your use case

Smallville7123 · Answer

modifying a class

diesieben07 · Answer

That is very vague and I can t really see the connection to modifying a parameter passed in

Smallville7123 · Answer

``` process a path a extension globalVariables kppMacroList fun process src String extension String macroTmp ArrayList lt Macro gt val destinationPreProcessed = File $src$ globalVariables preprocessedExtension $extension macro needs to be a modifiable value thus cannot be declared as a paramater directly var macro = macroTmp var index = macro size 1 if macro index fileName = null index++ println reallocating to $index macro = macro 0 realloc macro index + 1 macro index fileName = src substringAfterLast println registered macro definition for $ macro index fileName at index $index println processing $ macro index fileName gt $ destinationPreProcessed name destinationPreProcessed createNewFile val lex = Lexer fileToByteBuffer File src globalVariables tokensNewLine lex lex println lex currentLine is $ lex currentLine while lex currentLine = null val out = parse lex macro var input = lex currentLine as String if input input length 1 == n input = input dropLast 1 println ninput = $input println output = $out n n n n n n n n n n n n n n if globalVariables firstLine destinationPreProcessed writeText out + n globalVariables firstLine = false else destinationPreProcessed appendText out + n lex lex ```

ghedeon · Answer

gt modifying a parameter passed value is undefined behaviour It s so well defined to the extent of being forbidden

Smallville7123 · Answer

so what would be the correct way to accomplish this

Smallville7123 · Answer

which realloc is just ``` fun realloc m ArrayList lt Macro gt NewSize Int ArrayList lt Macro gt m add Macro m 0 size = NewSize return m ```

Smallville7123 · Answer

meaning to change it so it adds the correct number of new elements to the list instead of only 1 even if more than 1 is requested and delete if less than the current size is requested aswell

Smallville7123 · Answer

in ```macro = macro 0 realloc macro index + 1 ```

diesieben07 · Answer

`realloc` always returns the same object There is no need to return it or to re assign to `macro`

Smallville7123 · Answer

have no idea if reassigning it is nessisary or not

Smallville7123 · Answer

ok

Jonathan Mew · Answer

The intention behind the code is a little confusing it looks like the first element in your list of macros has some special role maintaining some kind of size for the whole list and the method to append a new macro to the list of macros is defined on the Macro object and specifically called on the first macro in the list But in any case the reason you don t need to reassign is that ``` here you pass in macro as the first argument and you assign the result back to macro macro = macro 0 realloc macro index + 1 here m is the original macro list fun realloc m ArrayList lt Macro gt NewSize Int ArrayList lt Macro gt m add Macro m 0 size = NewSize and it is returned here without having been modified it still refers to the original macros ArrayList return m so it currently reassigns macro to the value it already had ``` Hope that helps

Smallville7123 · Answer

ok