Array not displaying properly when passed to a function as pointer

Question

Given below is my main() function:

int main()
{
    int N = 4;
    int A[N][N] = {
        {1 , 0 , 0 , 0},
        {1 , 1 , 0 , 1},
        {0 , 1 , 0 , 0},
        {1 , 1 , 1 , 1}
    };
    for (int i = 0; i < N; ++i)
    {
        for (int j = 0; j < N; ++j)
            cout << A[i][j] << " ";
        cout << "\n";
    }
    cout << "\n";
    printSolution(N , *A);
    cout << "\n";
    return 0;
}

Here I had declared a 4x4 array with values. Given below is printSolution where I am passing a pointer to the array inside it.

void printSolution(int N , int *sol)
{
    for (int i = 0; i < N; ++i)
    {
        for (int j = 0; j < N; ++j)
            cout << *((sol + i) + j) << " ";
        cout << "\n";
    }
}

Given below is the output:

As it is visible in the output, the for loop inside the main function printed the array correctly, whereas the printSolution() function could not print it properly. Why is that so?

The question aside, this is C++ -- you should be using <array> or <vector>. — DevSolar
– DevSolar, Commented Jul 24, 2018 at 10:34
Ah. Another instructor teaching C first... he shouldn't be doing that... — DevSolar
– DevSolar, Commented Jul 24, 2018 at 10:36
int A[N][N] is not a valid declaration since N is not a constant expression. Many compilers allow it anyway as an unofficial extension called "variable length arrays", but those can bite you in tricky ways. Use constexpr int N = 4; (or const int N = 4;) instead. — aschepler
– aschepler, Commented Jul 24, 2018 at 11:11

bipll · Accepted Answer · 2018-07-24 10:37:25Z

4

*((sol + i) + j)

Say, for i = 2 and j = 2 this is merely *(sol + 4), the element in row 1 column 0 (exactly what is printed.)

You probably want *((sol + i * N) + j).

answered Jul 24, 2018 at 10:37

bipll

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6 Comments

Killzone Kid Over a year ago

Wouldn't this be an undefined behaviour? sol is a pointer to {1 , 0 , 0 , 0}, accessing other elements beyond this array is UB, no?

bipll Over a year ago

@KillzoneKid "A declaration of the form T a[N];, declares a as an array object that consists of N contiguously allocated objects of type T." Op passes *A as function's argument which is the first row of the array, other rows following contiguously.

Killzone Kid Over a year ago

@bipll the size N of the array sol points to is 4, it is not pointing to 1D array of int the size of N*4

Jarod42 Over a year ago

@KillzoneKid: indeed, pedantically sol[4] (A[0][4]) is UB. In practice, we read A[1][0].

Killzone Kid Over a year ago

@Jarod42 Would this be a better example of this kind of UB? ideone.com/dErrGU

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