Question

I am trying to write a simple C++ algorithm to solve sudoku. I am trying to pass address values between different functions but i get segmentation fault at runtime. (Needless to say, i am not quite experienced :))

The code does manage to pass the address of a[0] to main function and i can read values using pointers inside main. When i try to pass the address to solve function, it gives segmentation fault.

(Also as a secondary question, i can read values correctly in main, using cout << *(a+5) etc. correctly (commented out in main), but when i try to print all 81 values stored using a for loop, it gives out nonsense values (again, commented out in code). The code works with literals like *(a+3) or a[3], but does not when an int gets involved for(int i, whatever) cout << *(a+i);)

#include <iostream>
using namespace std;

int * get_input();
void solve(int *);

int main()
{
    int * a;
    a = get_input();
    //cout << *a << " " << *(a+1) << " " << *(a+2) << " " << *(a+3) << " " << *(a+4);
    //for (int i = 0 ; i < 81 ; i++) {if (i%9 == 0) cout << "\n"; cout << a[i] << " ";}
    solve(a);
    return(0);
}

int * get_input ()
{
    int a[81];
    getinput:
    for (int i = 0 ; i < 81 ; i++)  {a[i] = 0;}
    for (int i = 0 ; i < 81 ; i++)  {cin >> a[i];}
    print:
    for (int i = 0 ; i < 81 ; i++)
    {
        if (i%27 == 0){cout << "\n";}
        if (i%9 == 0) {cout << "\n";}
        if (i%3 == 0) {cout << "  " << a[i];}
        if (i%3 != 0) {cout << a[i];}
    }
    cout << "\n\nCheck:\n1- Fix\n2- Reset\n3- Confirm\n\n";
    int check = 0;
    cin >> check;
    if (check == 1)
    {   
        int input[3] = {-1, -1, -1};
        while (true)
        {
            cin >> input[0] >> input[1] >> input [2];
            if (input[1] == 0) goto print;
            a[(input[2]-1)+((input[1]-1)*9)] = input[0];
        }
    }
    if (check == 2) goto getinput;
    if (check == 3) return a;
}

void solve(int * a)
{
    bool matrix[9][9][9];
    for (int i = 0 ; i < 81 ; i++) {for (int j = 0 ; j < 9 ; j++) {matrix[(i-i%9)/9][i%9][j] = true;}}
    for (int i = 0 ; i < 81 ; i++)
    {
        if (a[i] == 0) continue;
        else
        {
            for (int j = 0 ; j < 9 ; i++)
            {
                matrix[(i-i%9)/9][j][a[i]] = false;
                matrix[j][i%9][a[i]] = false;
                matrix[((i-i%9)/9)-((i-i%9)/9)%3+j%3][i%9-(i%9)%3+(j-j%3)/3][a[i]] = false;
            }
        }
    }
    for (int i = 0 ; i < 9 ; i++)
    {
        for (int j = 0 ; j < 9 ; j++)
        {
            cout << matrix[i][j][1] << " ";
        }
        cout << "\n";
    }
}

Answer 1

Make the a array in your get_input() function static:

int a[81];

should be

static int a[81];

This works because the static keyword ensures that the allocated memory block (the array a ) will remain allocated after the function returns. Usually this is “because I’m not finished with it yet” (for example, you can count how many times your function is called this way), but it can also, legitimately, be used to ensure that the return value of a function survives the end of the function.

Marginally better would be to declare the array at the main level, and pass a pointer to it into the get_input() and solve() functions. That way you make it explicit in the code that the array will “live for the duration of the program” – and that’s usually good practice.