CODEWITHSUNDEEP
c++arrays
Varian Caesar
Varian Caesar

Reputation: 64

Array overflow c++ (at arr[0])?

Ok so last night I taught my sister some basic code how to read input, put it in array and print it to the screen. Here is the code :

int main() {
    int N;  // size of array

    cin >> N;
    int arr[N];

    for(int i = 1; i<= N; i++) {
        cin >> arr[i];
    }

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

    return 0;
}

And here is the result :

input :
4
2 3 4 1

output :
8
2
3
4

I don't have any clues why this can happen, I tried to check the code but it seems simple and already correct for me.

Upvotes: 1

Views: 200

Answers (4)

Houssam Badri
Houssam Badri

Reputation: 2509

An array of N elements has indexes from 0 to N-1 (so: N-1 - 0 + 1 = N elements, some maths.)

So fix this: for(int i = 1; i<= N; i++) with this : for(int i = 1; i< N; i++)

Upvotes: 0

Christos
Christos

Reputation: 53958

The first thing you have to look after is that dasblinkenlight has already mentioned in his answer, variable-length arrays are a language extension.

Then you should think that arrays are zero index based. What does this means? Let's consider an array of size 4 that is denoted as arr. Then it's elements can be accessed as arr[0], arr[1], arr[2], and arr[3]. So what happens when you run your first loop?

for(int i = 1; i<= N; i++) {
    cin >> arr[i];
}

You set the a value for arr[1], arr[2] to arr[N]. But the arr[N] points to a place in memory that is not associated with the array, since the array's size is N and all arrays are zero based. The last element of the array lives in here arr[n-1].

Unfortunately in C++ there isn't any check for the length of an array and you are allowed to do things like this. For instance, if you had tried this in C# or in Java then you would have got an Index Out Of Range Exception.

If you change your code to the following, then you will get the expected output:

for(int i = 0; i<N; i++) {
    cin >> arr[i];
}

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

Upvotes: 3

Vlad from Moscow
Vlad from Moscow

Reputation: 310950

First of all C++ does not support variable length arrays and if an array is declared as having N elements then the valid range of indices is [0, N-1].

You could either dynamically allocate an array of the required size yourself or use standard container std::vector<int>.

Using std::vector<int> you could for example write

#include <iostream>
#include <vector>

int main() 
{
    unsigned int n = 0;

    std::cout << "Enter number of elements: ";
    std::cin >> n;

    std::vector<int> v( n );

    std::cout << "Enter " << n << " elements: ";

    for ( int &x : v ) std::cin >> x;

    std::cout << "\nThe array is ";

    for ( int x : v ) std::cout << x << ' ';
    std::cout << std::endl;

    return 0;
}

The program output might look like

Enter number of elements: 10
Enter 10 elements: 0 1 2 3 4 5 6 7 8 9

The array is 0 1 2 3 4 5 6 7 8 9

Upvotes: 4

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726509

Your code is not valid C++, because variable-length arrays are a language extension.

With that part out of our way, consider what happens in your first loop's last step: you are accessing an element past the end of the array, which is undefined behavior.

Your second loop is correct, though, so all you need to do is correcting the header of the first loop to match the header of the second loop.

Upvotes: 2

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