Question

I’m trying to create a function that would dynamically allocate an array, sets the values of the elements, and returns the size of the array. The array variable is a pointer that is declared outside the function and passed as a parameter. Here is the code:

#include <cstdlib>  
#include <iostream>  
using namespace std;  

int doArray(int *arr) {  
    int sz = 10;  
    arr = (int*) malloc(sizeof(int) * sz);  

    for (int i=0; i<sz; i++) {  
        arr[i] = i * 5;  
    }  

    return sz;  
}  

int main(int argc, char *argv[]) {  

    int *arr = NULL;  
    int size = doArray(arr);  

    for (int i=0; i<size; i++) {  
        cout << arr[i] << endl;  
    }  

    return 0;  

}  

For some reason, the program terminates on the first iteration of the for loop in main()! Am I doing something wrong?

Answer 1

You’re passing in the array pointer by value; this means that when your doArray function returns, the value in arr in main is still NULL – the assignment inside doArray doesn’t change it.

If you want to change the value of arr (which is an int *), you need to pass in either a pointer or a reference to it; hence, your function signature will contain either (int *&arr) or (int **arr). If you pass it in as a ** you’ll also have to change the syntax inside the function from using arr to *arr (pointer-dereferencing), and you’ll call it like so: doArray(&arr).

Also, in C++ you should really be using new int[sz] instead of malloc.