Null pointer and pointer arithmetic

Question

Is there a way to stop the while loop below from iterating after surpassing 40? I am trying to replicate the linked list concept of iterating while NULL pointer is not found.

int main() {
    int* arr = new int[4]{ 10,20,30,40 };
    //for(int i=0; i<4; ++i) -- works fine
    while (arr) {
        cout << *(arr++) << endl;
    }
        
    delete[] arr; // Free allocated memory
    return 0;
}

Pointers don't become NULL when they reach the end of the array. C++ does not work this way. — Sam Varshavchik, Jul 29 '20 at 12:35
(arr+4) is NOT NULL. You may try to `std::cout<< (arr++)` and check what you get. — Louis Go, Jul 29 '20 at 12:35
The linked list concept applies to, well, linked lists. Dynamic array is a bad smell in modern C++. You want a `std::vector`. — sweenish, Jul 29 '20 at 12:37

score 1 · Answer 1 · answered Jul 29 '20 at 12:44

Because arr is placed in a contiguous memory, you will never get a NULL value of memory address AFTER arr.

You may try following code on online compiler.

#include <iostream>

int main()
{
    int* arr = new int[4]{ 10,20,30,40 };
    for(int i=0; i<4; ++i){
        std::cout << *(arr++) << std::endl;
        std::cout << arr << std::endl;
    }
    std::cout << "NULL is " << (int*)NULL; // NULL mostly stands for 0.
    return 0;
}

Output might be something like this:

10    
0x182de74    
20    
0x182de78    
30    
0x182de7c    
40    
0x182de80    
NULL is 0

Why does linkedlist works? Because linkedlist stores data in non-contiguous memory and next() would give you NULL as a sign of the end of a list.

Also you might need a fundamental book of C++.

Here's the booklist.

score 1 · Answer 2 · edited Jul 29 '20 at 15:25

1

Use a reserved value such as zero and append it to the end of the array, just like with an old C string. This is called a sentinel element.

int* arr = new int[4]{ 10,20,30,40,0 };
while (*arr) {
      ...

edited Jul 29 '20 at 15:25

Pablochaches

968
10
25

answered Jul 29 '20 at 13:05

eerorika · Accepted Answer · 2020-07-29T12:45:40.090

Is there are to stop the while loop below from iterating after surpassing 40

There are two ways to stop the loop: Cause the condition to become false, or jump out of the loop (break, goto, return, throw, etc.). Your loop doesn't do either.

Your condition is arr which is false only if the pointer points to null. You never assign null to arr, so it is never null.

I am trying to replicate the linked list concept

Linked list concepts do not generally apply to things that aren't linked lists. Arrays are not linked lists.

Amit G. · Answer 4 · 2020-08-21T03:04:46.627

int[4] is a C array. Instead, Use C++ std::array and its iterator:

#include <array>
#include <iostream>

int main() // Print all items
{
    std::array<int, 4> arr{ 10, 20, 30, 40 };

    for (auto i : arr)
        std::cout << i << std::endl;
}

Or:

#include <array>
#include <iostream>

int main() // Print until the 1st 0 item
{
    std::array<int, 6> arr{ 10, 20, 30, 40, 0, 0 };

    for (auto i : arr) {
        if (i == 0)
            break;
        std::cout << i << std::endl;
    }
}

Null pointer and pointer arithmetic

4 Answers4