C Program - Storing value in a pointer varaible

Question

I'm working on a code which ask user to type the number of rows and columns for 2 matrices and later ask user to type its elements.

After taking the user input for elements of matrices. The function matadd is called which receives the addresses of array variable a[20][20], b[20][20] containing elements of first and second matrix, address of variables m, n containing no. of rows and columns, address of array variable s[20][20] which is going to store the sum of two matrices.

matadd function code :

int matadd(int *(h)[20], int *(k)[20], int *u, int *l,int *(m)[20])
{
//VARIABLE-DECLARATION
int i = 0, j = 0;

for (i = 0; i < *u; i++)
{
    for (j = 0; j < *l; j++)
    {
        *(*(m + i) + j) = *(*(h + i) + j) + *(*(k + i) + j);
    }
}

printf("The Addition of above two given matrices is given below :\n");

printf("--           --\n");
for (i = 0; i < *u; i++)
{
     for (j = 0; j < *l; j++)
    {
        printf("|");
        for (j = 0; j < *l; j++)
        {
            printf("%4d", *(*(m + i) + j));
        }
        printf(" | ");
        printf("\n");
    }
}
printf("--           --\n");

return(m);

}

The problem comes at this line of code *(*(m + i) + j) = *(*(h + i) + j) + *(*(k + i) + j); where it has to assign the added value one by one inside pointer variable m.

But program stops there and give me error :

Exception thrown at 0x00C124B9 in mATRIX_pROTO.exe: 0xC0000005: Access violation reading location 0x00000001.

If there is a handler for this exception, the program may be safely continued.

Not able to find whats wrong with the code. I am using Visual Studio 2015 for debugging my code.

Any help would be appreciated.

Here's my code :

//CONSTANT-VARIABLE
#define MAX 1000

//USER-DEFINED FUCNTION
int matadd(int *h, int *k, int *u, int *l,int *(m)[20]);
char xgets(char *line, int size, FILE *stdi);
void header(void);

char xgets(char *line, int size, FILE *stdi) {
  /* read a line from the user */
 fgets(line, size, stdi);

  /* strip the \n if present */
  line[strcspn(line, "\n")] = '\0';
 return line;
}

void header(void)
{
printf("*-*-*-*-*MATRIX_ADD_PTR*-*-*-*-*");
printf("\n\n");
}

//PROGRAM STARTS HERE
int main(void)
{
 //FUCNTION CALL-OUT
header();

//VARIABLE-DECLARATION
int a[20][20] = { { 0 } }, b[20][20] = { { 0 } }, c[20][20] = { { 0 } },s[20][20] = { {0} };
int i = 0, j = 0;
int m = 0, n = 0, q = 0, w = 0;
int line[MAX] = { 0 };

printf("Type Same Number of ROWs And COLUMNs for Both Matrx \n\n");

printf("Enter the No. of ROWs needed in First Matrix : ");
xgets(line, sizeof(line), stdin);
sscanf_s(line, "%d", &m);
printf("\n");

printf("Enter the No. of COLUMNs needed in First Matrix : ");
xgets(line, sizeof(line), stdin);
sscanf_s(line, "%d", &n);
printf("\n");

printf("Enter the Elements for 1st ROW then 2nd and So on for First Matrix : \n");
for (i = 0; i < m; i++)
{
    for (j = 0; j < n; j++)
    {
        xgets(line, sizeof(line), stdin);
        sscanf_s(line, "%d", &a[i][j]);
    }
}

printf("\n");

printf("Enter the No. of ROWs needed in Second Matrix : ");
xgets(line, sizeof(line), stdin);
sscanf_s(line, "%d", &q);
printf("\n");

printf("Enter the No. of COLUMNs needed in Second Matrix : ");
xgets(line, sizeof(line), stdin);
sscanf_s(line, "%d", &w);
printf("\n");

printf("Enter the Elements for 1st ROW then 2nd and So on for Second Matrix : \n");
for (i = 0; i < q; i++)
{
    for (j = 0; j < w; j++)
    {
        xgets(line, sizeof(line), stdin);
        sscanf_s(line, "%d", &b[i][j]);
    }
}

//OUTPUT OF ENTERED MATRIX
printf("First Matrix : \n\n");

printf("--        --\n");
for (i = 0; i < m; i++)
{
    printf("|");
    for (j = 0; j < n; j++)
    {
        printf("%3d", a[i][j]);
    }
    printf(" | ");
    printf("\n");
}
printf("--        --\n\n\n");

printf("Second Matrix : \n\n");

printf("--        --\n");
for (i = 0;i < q;i++)
{
    printf("|");
    for (j = 0; j < q; j++)
    {
        printf("%3d", b[i][j]);
    }
    printf(" | ");
    printf("\n");
}
printf("--        --\n\n\n");

//FUCNTION CALL-OUT
matadd(a, b, &m, &n, s);

//TERMINAL-PAUSE
system("pause");
}


int matadd(int *(h)[20], int *(k)[20], int *u, int *l,int *(m)[20])
{
    //VARIABLE-DECLARATION
    int i = 0, j = 0;

    for (i = 0; i < *u; i++)
{
        for (j = 0; j < *l; j++)
     {
        *(*(m + i) + j) = *(*(h + i) + j) + *(*(k + i) + j);
     }
 }

printf("The Addition of above two given matrices is given below :\n");

printf("--           --\n");
for (i = 0; i < *u; i++)
{
    for (j = 0; j < *l; j++)
     {
        printf("|");
        for (j = 0; j < *l; j++)
         {
            printf("%4d", *(*(m + i) + j));
         }
        printf(" | ");
        printf("\n");
    }
}
printf("--           --\n");

return(m);

}

Answer 1

Looking at your addition code, we see this pattern three times:

*(*(h + i) + j)

The above expression should be of type int in order to make sense in the context of element wise addition. That means that

(*(h + i) + j)

should be of type int *, I.e. a pointer to an integer. This also means that j should be meaningfully convertible to size_t (it is) and that

*(h + i)

should also be of type int *. Then

(h + i)

should be a pointer to the above type, thus a int **. similar to above i should be convertible to size_t (which it is) and h should've the same type.

More directly speaking, it should be a pointer to (an array of) pointer(s) to (an array of) int(s).

Let's see what you have in main:

int a[20][20];

That's an array of arrays, thus except for indexing similar to

int a[20*20];

in that it's stored in contiguous memory. That's a different representation as above!

You need to decide on one of these representations.

I try to visualise this using a smaller, 2 on 2 situation (I'm on a mobile, thus this will be hard enough this way):

data:
1 2
3 4

Your function is expecting your data like that:

@some_address:
  address_one <--- h points to this
  address_two

@address_one:
  1
  2

@address_two:
  3
  4

But you have it stored like that:

@some_address:
  1 <--- a decays to a pointer pointing to here upon use
  2
  3
  4

Thus, your function uses the values 1 and 2 as memory addresses.

Note that I'm only addressing the (IMHO) main issue here, there are a lot more things to fix, as the example compiler messages in the other answer show.

Answer 2

You cannot just ignore all the warnings issued by the Visual Studio Compiler. After adding the mentioned includes stdlib.h, stdio.h and process.h and adding the switch /TC to the compiler call, the following warnings

warning C4047: 'function' : 'int *' differs in levels of indirection from 'int [20][20]'

warning C4024: 'matadd' : different types for formal and actual parameter 1

at the line

matadd(a, b, &m, &n, s);

clearly indicate that something is going wrong in your code. These warnings are issued because your function prototype expect a int * where you have given a int [20][20] in the call. Nevertheless, the function prototype does not match your definition of matadd. If I change the prototype to

int matadd(int *(h)[20], int *(k)[20], int *u, int *l,int *(m)[20]);

the compiler issues the following warnings instead:

warning C4047: 'function' : 'int **' differs in levels of indirection from 'int [20][20]'

warning C4024: 'matadd' : different types for formal and actual parameter 1

You can fix it by the changing the prototype and definition to:

int matadd(int (*h)[20], int (*k)[20], int *u, int *l,int (*m)[20]);

EDIT: The type int (*h)[20] actually represents a pointer to an array with 20 elements. If you add 1 to the pointer and dereference it with (*(h+1)), then you get the next 20 elements after the first 20 elements pointed to by (*h), and so on. You can also express it with the declaration int h[][20] which shows better the intended usage as a 2D array. Note, only the size of the first dimension can be left undefined. (There are even more possibilities as discussed here and here, but Visual Studio 2010 do not support these.)

EDIT: You can implement the matadd function without any explicit pointer arithmetic. This is a stripped down example:

#include <stdio.h>

// specifying the size of the first dimension is optional
void matadd(int h[][20], int k[][20], int u, int l, int m[][20])
{
    int i = 0, j = 0;

    for (i = 0; i < u; i++)
    {
        for (j = 0; j < l; j++)
        {
            m[i][j] = h[i][j] + k[i][j];
        }
    }
}

int main(void)
{
    int i, j, m, n;
    int a[20][20] = { { 0 } }, b[20][20] = { { 0 } }, s[20][20] = { {0} };

    m = 2; n = 2;
    a[0][0] = 1; a[0][1] = 2; a[1][0] = 3; a[1][1] = 4;
    b[0][0] = 5; b[0][1] = 6; b[1][0] = 7; b[1][1] = 8;

    matadd(a, b, m, n, s);

    for (i = 0; i < m; i++)
    {
        for (j = 0; j < n; j++)
        {
            printf("%4d | ", s[i][j]);
        }
        printf("\n");
    }
}

Your code even needs more fixes as indicated by compiler warnings;

warning C4013: 'strcspn' undefined; assuming extern returning int

because you forgot to include string.h.

warning C4047: 'function' : 'char' differs in levels of indirection from 'char *'

at return line; and return(m);. Here, you are converting a pointer to a character. The program does not segfault here because you happily did not use the return value.

warning C4133: 'function' : incompatible types - from 'int [1000]' to 'char *' warning C4133: 'function' : incompatible types - from 'int [1000]' to 'const char *'

at the calls of xgets and sscanf_s. This type conversion is also invalid. The program does not segfault, because the int [1000] is happily larger than a corresponding char [1000].

---

My previous long comment which was the starter of the discussion in the comments can be found here.