Question

int main(){

    int right[2][3] = {
    {1,4,6}, {2,7,5}
    }

    ....


    calc(right);
}

int calc(int ** right){
    printf("%i", right[0][0]);
}

I calc function that calculate some numbers based on a matrix, but I dont’ know why i get seg fault when I access the variable right within the calc function. does any body know the solution?

edit:
right now that is all it’s doing at calc function. I have some calc stuff but it’s all commented out trying to figure out how to access this variable.

Answer 1

Two-dimensional arrays in C don’t work the way you think they do. (Don’t worry, you’re not alone — this is a common misconception.)

The assumption implicit in the code is that right is an array of int * pointers, each of which points to an array of int. It could be done this way — and, confusingly, the syntax for accessing such an array would be the same, which is probably what causes this misconception.

What C actually does is to make right an array of 12 ints, layed out contiguously in memory. An array access like this

a=right[i][j];

is effectively equivalent to this:

int *right_one_dimensional=(int *)right;
a=right[i*3 + j];

To pass your array to the calc function, you need to do this:

int calc(int *right, size_t d){
    // For example
    a=right[i*d + j];
}

and then call it like this:

int right[2][3] = {
    {1,4,6}, {2,7,5}
};

calc(&right[0][0], 3);

Edit: For more background on this, the question linked to in Binary Worrier’s comment is definitely worth looking at.