How to execute a unique shuffle using two for loops - c++

Question

I'm trying to write a function called func1 that can shuffle a given deck of cards (an array) uniquely, given two other decks, the starting deck and the resulting deck. For example, a deck of cards [1, 2, 3, 4, 5] is shuffled and produces another deck of cards [4, 5, 2, 1, 3]. I want to complete this same shuffle (putting the card in slot 0 into slot 3, the card in slot 1 into slot 2, etc), but this time on another deck of cards [2, 3, 1, 5, 4]. It should print out [5, 4, 3, 2, 1] if I wrote the code correctly. Although it runs the program, it only prints the first "card" correctly, and the rest are really big numbers. What did I do wrong? Am I using the correct approach, or should I rethink my design?
#include <iostream>
using namespace std;
int * func1(int *deck) // A unique shuffle given two decks, a starting and ending deck (in this case, "start" and "shuff1")
{
int start[5] = { 1,2,3,4,5 }; // Starting deck
int shuff1[5] = { 4,5,2,1,3 }; // Resulting deck after shuffle. This is the specific type of shuffle that we are copying
int finish[5] = {}; // The deck that we are returning
for (int i = 0; i < 5; i++) // Looks at a specific spot (i) in the start deck...
{
for (int j = 0; j < 5; j++) // Looks through all the spots (j) in the shuff1 deck...
{
if (start[i] == shuff1[j]) // And if the cards themselves are identical, then it takes the ith card
{ // in the given deck and puts it in the jth spot in the finish deck
int temp = deck[i];
finish[j] = temp;
j = 5;
}
}
}
return finish;
}
int main()
{
int test[5] = { 2,3,1,5,4 }; // Given deck
int* shuff2 = func1(test); // Calls a specifc shuffle and sets it equal to a deck called shuff2
for (int i = 0; i < 5; i++)
{
cout << shuff2[i] << endl; // Prints shuff2
}
return 0;
}

Answer 1

Your func1 returns a pointer to finish, which is a variable local to the function. When control passes out of the function, that variable passes out of scope-- you could say it expires. So the pointer points to a spot in the wilderness of invalid memory, and dereferencing the pointer (e.g. by shuff2[i]) causes undefined behavior, which means strange numbers if you're lucky.
One solution is to have func1 construct finish on the heap. Another is to have the function accept a pointer to the array which it should fill with shuffled values; the calling code (e.g. main) would be responsible for providing a pointer to a valid array.

Answer 2

Your main problem is that you are returning a pointer to a variable allocated on the stack, which you can't do. Pass a pointer to "finish" as a second argument instead.
I'm not sure if I am following the way you described the proposed algorithm very well, but I think you would want to create an array of offsets, like this:
void func1(int *deck, int *finish) {
int offsets[5] = {3, 1, 2, -3, -3};
for (int i = 0; i < 5; ++i) {
finish[i + offsets[i]] = deck[i];
}
}
or put the actual destination indexes into the array:
void func1(int *deck, int *finish) {
int desti[5] = {3, 2, 4, 0, 1};
for (int i = 0; i < 5; ++i) {
finish[desti[i]] = deck[i];
}
}
Though you did specifically say "using two for loops" so maybe I'm misunderstanding the way you described the algorithm.