Pass by value for recursion?

Question

I am trying to pass a parameter "by value". I have tried making a deep copy of the parameter that is passed recursively in order to prevent any changes from circling back to the parent function calls.

Here is a snippet of code that tries to generate the array of all possible parentheses.

def generateParenthesis(n):
    #Iterate for each move.
    M = 2 * n
    retArray = []
    def recHelper(numMoves, perm, stack):
        print("Function call: ", numMoves, perm, stack)
        newPerm = copy.deepcopy(perm)
        newStack = stack.copy()
        #Base case, no moves
        if (numMoves == 0):
            retArray.append(newPerm)
            return

        #Case when left move is valid
        if (numMoves != len(newStack)):
            #Apply the left move. Pass it recursively
            newPerm +='('
            #Update the stack accordingly
            newStack.append('(')
            #Decrease numMoves
            newNumMoves = numMoves - 1
            #Call it recursively
            recHelper(newNumMoves, newPerm, newStack)
        #Case when right move is valid
        if len(newStack) != 0:
            #Apply the right move. Pass it recursively
            newPerm +=')'
            #Update the stack accordingly, delete the top, last elm
            newStack.pop()
            #Decrease numMoves
            newNumMoves = numMoves - 1
            #Call it recursively
            recHelper(newNumMoves, newPerm, newStack)
        #done
        return
    recHelper(M, "", [])
    return retArray

Unfortunately, calling generateParenthesis(1) returns ['()','()(', '()()'] and not ['()'].

Answer 1

Use + operator to add to a list rather than .append() behavior to best emulate pass by value behavior.

Python officially abides by pass-by-object-reference. In your case, when a list stack or perm passed and modified in the child function, the parent function's stack or perm will see the updated variable.