Passing in array to a function

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.

You need to post the code for calc function for any body to help. — Naveen
– Naveen, Commented Sep 8, 2010 at 7:37
You probably try to access the whole array but you only initialize right[0] and [1] IMO. — InsertNickHere
– InsertNickHere, Commented Sep 8, 2010 at 7:39
can you paste the code of calc as well?and why are you using int ** right?? — Anil Vishnoi
– Anil Vishnoi, Commented Sep 8, 2010 at 7:39
possible duplicate of many questions, including Passing arrays and matrices to functions as pointers and pointers to pointers in C — Binary Worrier
– Binary Worrier, Commented Sep 8, 2010 at 7:46

Martin B · Accepted Answer · 2010-09-08 07:49:07Z

7

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.

answered Sep 8, 2010 at 7:49

Martin B

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1 Comment

nos Over a year ago

You could also specify calc as int calc(int right[][3]) and call it by simple doing calc(right);

Bart van Ingen Schenau · Accepted Answer · 2010-09-08 08:01:27Z

Although a one-dimensional array is automatically converted to a pointer, the same does not hold for a multi-dimensional array and multi-level pointers.

If you change the order of the calc and main functions (or if you provide a prototype for calc before main), you will get a complaint from the compiler that it can convert right to the type int**.

The reason is that right is declared as an "array of 4 arrays of 3 int". This can be automatically converted to "pointer to array of 3 int" (int (*)[3]), but that is where the conversions stop. calc on the other hand expects a "pointer to a pointer to int", which is a completely different beast from a "pointer to array of 3 int".

There are two possible solutions:

Change calc to accept a pointer to an array (or array of arrays):

   int calc(int right[][3])

Change right to be a pointer to a pointer:

    int temp_array[4][3];
    int* temp_array2[4] = { temp_array[0], temp_array[1], temp_array[2], temp_array[3] };
    int** right = temp_array2;

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