CODEWITHSUNDEEP
c++arrayspointerspointer-arithmetic
motiur
motiur

Reputation: 1680

Null pointer and pointer arithmetic

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;
}

Upvotes: 0

Views: 169

Answers (4)

Amit G.
Amit G.

Reputation: 2684

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;
    }
}

Upvotes: 0

user1196549
user1196549

Reputation:

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) {
      ...

Upvotes: 1

eerorika
eerorika

Reputation: 238421

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.

Upvotes: 0

Louis Go
Louis Go

Reputation: 2588

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.

Upvotes: 1

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