CODEWITHSUNDEEP
c++arrayspointers
dmarzio
dmarzio

Reputation: 355

Pointer to an array isn't working as intended

I'm trying to figure out an implementation of an array-based queue in c++. At this point I'm just trying to initialize an array with a size dependent on what the user inputs. I have a Queue class as follows:

Queue.h: #include 

using namespace std;

class Queue {

    private:
        int *ptrtoArray;
        int arraysize;
        int data;
    public:
        Queue(int size);
        Queue();
        void print();


};

Queue.cpp:

#include "Queue.h"

Queue::Queue() {

}
Queue::Queue(int size) {
    int theQueue[size];
    arraysize = size;
    ptrtoArray = &theQueue[0];

    for (int i=0;i<size;i++) {
        theQueue[i] = -1;
    }
    cout << "(";
    for(int i=0;i<size;i++) {

        cout << ptrtoArray[i] << ",";
    }
    cout << ")" << endl;
}
void Queue::print() {
    cout << "(";
    for(int i=0;i<arraysize;i++) {
        cout << ptrtoArray[i] << ",";
    }
    cout << ")" << endl;
}

main.cpp:

#include <iostream>
#include "Queue.h"

using namespace std;

const string prompt = "queue> "; // store prompt

Queue *queue;

void options(){
    string input; // store command entered by user
    cout << prompt; // print prompt
    cin >> input; // read in command entered by user
    int queuesize,num;
    while(input != "quit") { // repeat loop as long as user does not enter quit
        if (input == "new") { // user entered ad
            cin >> queuesize;
            queue = new Queue(queuesize);
        } else if (input == "enqueue"){ //user entered remove
        } else if (input == "dequeue"){ //user entered remove
        } else if (input == "print"){ //user entered quit
            queue->print();
        } else { // no valid command has been selected
            cout << "Error! Enter add, remove, print or quit." << endl;
        }
        cout << prompt; // repeat prompt
        cin >> input; // read new input
    }
}//options

/**
 * Main function of the program. Calls "options", which handles I/O. 
 */ 
int main() {
    options();
    return 0;
}

When the code is executed everything is fine with the code inside the constructor, but the print function is giving some strange results.

queue> new 5
(-1,-1,-1,-1,-1,)
queue> print
(134516760,134525184,-1081449960,4717630,134525184,)
queue>

So, at this point I just want the print() function to show me that the array contains just -1 in every element. My knowledge of pointers is very limited, so if you could help me realize what I'm doing wrong that would be great!

Upvotes: 0

Views: 150

Answers (3)

user995502
user995502

Reputation: 

int theQueue[size];
arraysize = size;
ptrtoArray = &theQueue[0];

Allocates static array theQueue which will be freed when the constructor returns. You should instead allocate dynamic array with new

arraysize = size;
ptrtoArray = new int[size];

Don't forget to delete it in you destructor (or earlier if you don't need it any more).

EDIT

Static

int theQueue[size];

Will be allocated on the stack. Which means when your function (where the declaration is in) returns the memory is not accessible anymore. It can for example be used by the next function to hold another array.

Advantages faster, you don't have to explicitly free it so it avoids leaks.

Dynamic 

ptrtoArray = new int[size];

is allocated on heap. Which means even if you go out of scope it will still be owned by you (unless you lose the pointer to it. In which case you are doomed). If you don't need it anymore you can free it by using

delete[] ptrtoArray;

Advantages can have dynamic sizes. Are available out of scope.

Upvotes: 2

amrith92
amrith92

Reputation: 220

That's actually expected. See, the ptrtoArray member of class Queue points to a function-scope variable, which will not be available throughout the lifetime of the program. Here, that's your constructor. So ptrtoArray will be pointing to a completely random memory location, i.e. it's what is termed as a "wild-pointer"

See this: https://stackoverflow.com/a/11137525/1919302 for a better idea about the difference between Scope and Lifetime of a variable and why your output is the way it is.

Upvotes: 0

Rob G
Rob G

Reputation: 3526

You're declaring your array in local scope, then losing it as it falls out of scope, yet the pointer still points to its address in memory, which who knows what will become of it, as you've found out. Try:

ptrToArray = (int *) malloc(sizeof(int) * size);

or the new keyword:

ptrToArray = new int[size];

Upvotes: 3

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