Segmentation fault when running C program (putchar() and multidimensional array pointers)

Question

I'm trying to run the following code on C:

#include "ex1_1.h"

void path(char **adj_mat, int u, int v)
{
        printf("test\n");
        char temp = *adj_mat[1];
        putchar(temp);
}

int main()
{
    int u = 5;
    int v = 5;
    char mat[5][5]={
    {'0', '1', '1', '1', '0'},
    {'0', '0', '0', '0', '1'},
    {'0', '0', '0', '0', '0'},
    {'0', '0', '0', '0', '0'},
    {'0', '0', '0', '0', '0'}
    };
    char** adj_mat = (char**)&mat;
    printf("Mtest\n");
    path(adj_mat, u, v);
    return 1;
}

and I'm getting "Segmentation fault".

I don't know why, I'm guessing that it's related to how I'm using putchar() with multidimensional array pointers.

Thanks.

Answer 1

A char[5][5] is not a char ** .

Declare your path function as one of these (they are equivalent):

void path(char (*adj_mat)[5], int u, int v)
void path(char adj_mat[5][5], int u, int v)

and use it like

path(mat, u, v);

---

Update: Now to the reason for the segfault.

mat is a 5-array of 5-arrays of char , which will "decay" into a pointer to 5-arrays of char when used as an expression. This means that mat can be used as if it is a pointer to a 5-array. In general a n-array of type T will decay into a pointer of type T (note that T does not decay!).

So note you use mat as an expression when you pass it to your function pass() . In C it is not possible to pass an array as-is to a function, but it can only "receive" pointers.

So what's the difference between a pointer to pointer of char and a pointer to a 5-array of char ? Their type. And their sizes.

Suppose you have a pointer to a pointer to a char - let's call it p and we have a pointer to a 5-array q . In short: char **p and char (*q)[5] .

When you write p[i] , this will be equivalent to *(p+i) . Here you have pointer arithmetic. p+i has the value (which is an adress) p plus i*sizeof(char *) because p points to a char * .

When you look at q , q+i would have the value q plus i*sizeof(char[5]) .

These will almost always be different!

---

Update (real answer now): It's even worse in your case as you are "forcing" char (*q)[5] "to be" a char p** (via your invalid typecast), so in your call

char temp = *adj_mat[1];

adj_mat[1] will look at the 2nd row of your matrix an try to interpret its value as a pointer (but it is a 5-array!), so when you then dereference via *adj_mat[1] you will land somewhere in nirvana - segfault!

Answer 2

The conversion (char**)&mat is undefined behavior. &mat is char(*)[5][5] (In words, pointer to array of 5 arrays of 5 chars), and you can't convert it to pointer to pointer to char.

I would write path like (assuming C99):

void path(char *arr,int u,int v) {
 char (*adj_mat)[v]=arr;
 ...
}

Answer 3

Change the line

char temp = *adj_mat[1];

to

char temp = adj_mat[1];

The problem is that in

*adj_mat[1];

you try to dereference a pointer to an invalid address, namely the ASCII code of '1' (I suppose).