I have a function in C that one of its arguments is 3D array.
when I'm debugging its seems the VS see it as a 2D array, why is it?
this is a program that helps to order goods in containers (boxes).
this specific function is called when its known the this spot is ok to put' and it put it on the 3D matrix.the code:
void put_box(unsigned char space[241][47][48], int x, int y, int z, BOX box, int box_number) {
int i, j, k;
int top_stop;
//cheking if to mark all the hight
if (box.top)
top_stop = 48;
else
top_stop = box.height / 5;
// mark the box
for (i = x; i < x + (box.length / 5); i++)
for (j = y; j < y + (box.width / 5); j++)
for (k = z; k < top_stop; k++)
space[i][j][k] = (unsigned char)box_number;
}/*void put_box*/
I expect to see on the locals window the 3D array but I see only 2.
and its effect the writing to the matrix.
When you pass arrays to functions, they are passed as pointers. More specifically pointer to the arrays first element.
So for the argument declaration unsigned char space[241][47][48] the compiler really sees it as unsigned char (*space)[47][48].
Probably because the first array dimension is not part of the func prototype, so the debugger is likely to be looking at the first slice only.
void put_box(unsigned char space[][47][48], <snip> )
I'm guessing that if however you stick a breakpoint in this method, you'd see the full array (but you'll only see one slice when you step into put_box)
int main()
{
unsigned char space[99][47][48];
put_box(space, <snip>);
return 0;
}
Related
I dynamically allocated memory for 3D array of pointers. My question is how many pointers do I have? I mean, do I have X·Y number of pointers pointing to an array of double or X·Y·Z pointers pointing to a double element or is there another variant?
double*** arr;
arr = (double***)calloc(X, sizeof(double));
for (int i = 0; i < X; ++i) {
*(arr + i) = (double**)calloc(Y, sizeof(double));
for (int k = 0; k < Y; ++k) {
*(*(arr+i) + k) = (double*)calloc(Z, sizeof(double));
}
}
The code you apparently intended to write would start:
double ***arr = calloc(X, sizeof *arr);
Notes:
Here we define one pointer, arr, and set it to point to memory provided by calloc.
Using sizeof (double) with this is wrong; arr is going to point to things of type double **, so we want the size of that. The sizeof operator accepts either types in parentheses or objects. So we can write sizeof *arr to mean “the size of a thing that arr will point to”. This always gets the right size for whatever arr points to; we never have to figure out the type.
There is no need to use calloc if we are going to assign values to all of the elements. We can use just double ***arr = malloc(X * sizeof *arr);.
In C, there is no need to cast the return value of calloc or malloc. Its type is void *, and the compiler will automatically convert that to whatever pointer type we assign it to. If the compiler complains, you are probably using a C++ compiler, not a C compiler, and the rules are different.
You should check the return value from calloc or malloc in case not enough memory was available. For brevity, I omit showing the code for that.
Then the code would continue:
for (ptrdiff_t i = 0; i < X; ++i)
{
arr[i] = calloc(Y, sizeof *arr[i]);
…
}
Notes:
Here we assign values to the X pointers that arr points to.
ptrdiff_t is defined in stddef.h. You should generally use it for array indices, unless there is a reason to use another type.
arr[i] is equivalent to *(arr + i) but is generally easier for humans to read and think about.
As before sizeof *arr[i] automatically gives us the right size for the pointer we are setting, arr[i].
Finally, the … in there is:
for (ptrdiff_t k = 0; k < Y; ++k)
arr[i][k] = calloc(Z, sizeof *arr[i][k]);
Notes:
Here we assign values to the Y pointers that arr[i] points to, and this loop is inside the loop on i that executes X times, so this code assigns XY pointers in total.
So the answer to your question is we have 1 + X + XY pointers.
Nobody producing good commercial code uses this. Using pointers-to-pointers-to-pointers is bad for the hardware (meaning inefficient in performance) because the processor generally cannot predict where a pointer points to until it fetches it. Accessing some member of your array, arr[i][j][k], requires loading three pointers from memory.
In most C implementations, you can simply allocate a three-dimensional array:
double (*arr)[Y][Z] = calloc(X, sizeof *arr);
With this, when you access arr[i][j][k], the compiler will calculate the address (as, in effect, arr + (i*Y + j)*Z + k). Although that involves several multiplications and additions, they are fairly simple for modern processors and are likely as fast or faster than fetching pointers from memory and they leave the processor’s load-store unit free to fetch the actual array data. Also, when you are using the same i and/or j repeatedly, the compiler likely generates code that keeps i*Y and/or (i*Y + j)*Z around for multiple uses without recalculating them.
Well, short answer is: it is not known.
As a classic example, keep in mind the main() prototype
int main( int argc, char** argv);
argc keeps the number of pointers. Without it we do not know how many they are. The system builds the array argv, gently updates argc with the value and then launches the program.
Back to your array
double*** arr;
All you know is that
arr is a pointer.
*arr is double**, also a pointer
**arr is double*, also a pointer
***arr is a double.
What you will get in code depends on how you build this. A common way if you need an array of arrays and things like that is to mimic the system and use a few unsigned and wrap them all with the pointers into a struct like
typedef struct
{
int n_planes;
int n_rows;
int n_columns;
double*** array;
} Cube;
A CSV file for example is char ** **, a sheet workbook is char ** ** ** and it is a bit scary, but works. For each ** a counter is needed, as said above about main()
A C example
The code below uses arr, declared as double***, to
store a pointer to a pointer to a pointer to a double
prints the value using the 3 pointers
then uses arr again to build a cube of X*Y*Z doubles, using a bit of math to set values to 9XY9.Z9
the program uses 2, 3 and 4 for a total of 24 values
lists the full array
list the first and the very last element, arr[0][0][0] and arr[X-1][Y-1][Z-1]
frees the whole thing in reverse order
The code
#include <stdio.h>
#include <stdlib.h>
typedef struct
{
int n_planes;
int n_rows;
int n_columns;
double*** array;
} Cube;
int print_array(double***, int, int, int);
int main(void)
{
double sample = 20.21;
double* pDouble = &sample;
double** ppDouble = &pDouble;
double*** arr = &ppDouble;
printf("***arr is %.2ff\n", ***arr);
printf("original double is %.2ff\n", sample);
printf("*pDouble is %.2ff\n", *pDouble);
printf("**ppDouble is %.2ff\n", **ppDouble);
// but we can build a cube of XxYxZ doubles for arr
int X = 2;
int Y = 3;
int Z = 4; // 24 elements
arr = (double***)malloc(X * sizeof(double**));
// now each arr[i] must point to an array of double**
for (int i = 0; i < X; i += 1)
{
arr[i] = (double**)malloc(Y * sizeof(double*));
for (int j = 0; j < Y; j += 1)
{
arr[i][j] = (double*)malloc(Z * sizeof(double));
for (int k = 0; k < Z; k += 1)
{
arr[i][j][k] = (100. * i) + (10. * j) + (.1 * k) + 9009.09;
}
}
}
print_array(arr, X, Y, Z);
printf("\n\
Test: first element is arr[%d][%d[%d] = %6.2f (9XY9.Z9)\n\
last element is arr[%d][%d[%d] = %6.2f (9XY9.Z9)\n",
0, 0, 0, arr[0][0][0],
(X-1), (Y-1), (Z-1), arr[X-1][Y-1][Z-1]
);
// now to free this monster
for (int x = 0; x < X; x += 1)
{
for (int y = 0; y < Y; y += 1)
{
free(arr[x][y]); // the Z rows
}
free(arr[x]); // the plane Y
}
free(arr); // the initial pointer;
return 0;
}; // main()
int print_array(double*** block, int I, int J, int K)
{
for (int a = 0; a < I; a += 1)
{
printf("\nPlane %d\n\n", a);
for (int b = 0; b < J; b += 1)
{
for (int c = 0; c < K; c += 1)
{
printf("%6.2f ", block[a][b][c]);
}
printf("\n");
}
}
return 0;
}; // print_array()
The output
***arr is 20.21f
original double is 20.21f
*pDouble is 20.21f
**ppDouble is 20.21f
Plane 0
9009.09 9009.19 9009.29 9009.39
9019.09 9019.19 9019.29 9019.39
9029.09 9029.19 9029.29 9029.39
Plane 1
9109.09 9109.19 9109.29 9109.39
9119.09 9119.19 9119.29 9119.39
9129.09 9129.19 9129.29 9129.39
Test: first element is arr[0][0[0] = 9009.09 (9XY9.Z9)
last element is arr[1][2[3] = 9129.39 (9XY9.Z9)
I have a code from Mathlab, where all matrix operations are done by a couple of symbols. By translating it into C I faced a problem that for every size of matrix I have to create a special function. It's a big code, i will not place it all here but will try to explain how it works.
I also have a big loop where a lot of matrix operations are going on. Functions which are operating with matrices should take matrices as income and store results in temporary matrices for upcoming operations. In fact i know the size of matrices but i also want to make the functions as universal as possible. In oder to reduce code size and save my time.
For example, matrix transposition operation of 2x4 and 4x4 matrices:
void A_matrix_transposition (float transposed_matrix[4][2], float matrix[2][4], int rows_in_matrix, int columnes_in_matrix);
void B_matrix_transposition (float transposed_matrix[4][4], float matrix[4][4], int rows_in_matrix, int columnes_in_matrix);
int main() {
float transposed_matrix_A[4][2]; //temporary matrices
float transposed_matrix_B[4][4];
float input_matrix_A[2][4], input_matrix_B[4][4]; //input matrices with numbers
A_matrix_transposition (transposed_matrix_A, input_matrix_A, 2, 4);
B_matrix_transposition (transposed_matrix_B, input_matrix_B, 4, 4);
// after calling the functions i want to use temporary matrices again. How do I pass them to other functions if i dont know their size, in general?
}
void A_matrix_transposition (float transposed_matrix[4][2], float matrix[2][4], int rows_in_matrix, int columnes_in_matrix)
{ static int i,j;
for(i = 0; i < rows_in_matrix; ++i) {
for(j = 0; j < columnes_in_matrix; ++j)
{ transposed_matrix[j][i] = matrix[i][j];
}
}
}
void B_matrix_transposition (float transposed_matrix[4][4], float matrix[4][4], int rows_in_matrix, int columnes_in_matrix)
{ static int i,j;
for(i = 0; i < rows_in_matrix; ++i) {
for(j = 0; j < columnes_in_matrix; ++j)
{ transposed_matrix[j][i] = matrix[i][j];
}
}
}
The operation is simple, but the code is massive already because of 2 different functions, but it will be a slow disaster if I continue like this.
How do i create one function for transposing to operate matrices of different sizes?
I suppose it can be done with pointers, but I don't know how.
I'm looking for a realy general answer to understand how to tune up the "comunication" between functions and temporary matrices, best with an example. Thank you all in advance for the information and help.
There are different way you can achieve this in c from not so good to good solutions.
If you know what the maximum size of the matrices would be you can create a matrix big enough to accommodate that size and work on it. If it is lesser than that - no problem write custom operations only considering that small sub-matrix rather than the whole one.
Another solution is to - create a data structure to hold the matrix this may vary from jagged array creation which can be done using the attribute that is stored in the structure itself. For example: number of rows and column information will be stored in the structure itself. Jagged array gives you the benefit that now you can allocate de-allocate memory - giving you a better control over the form - order of the matrices. This is better in that - now you can pass two matrices of different sizes and the functions all see that structure which contain the actual matrix and work on it. (wrapped I would say).
By Structure I meant something like
struct matrix{
int ** mat;
int row;
int col;
}
If your C implementation supports variable length arrays, then you can accomplish this with:
void matrix_transposition(size_t M, size_t N,
float Destination[M][N], const float Source[N][M])
{
for (size_t m = 0; m < M; ++m)
for (size_t n = 0; n < N; ++n)
Destination[m][n] = Source[n][m];
}
If your C implementation does not support variable length arrays, but does allow pointers to arrays to be cast to pointers to elements and used to access a two-dimensional array as if it were one-dimensional (this is not standard C but may be supported by a compiler), you can use:
void matrix_transposition(size_t M, size_t N,
float *Destination, const float *Source)
{
for (size_t m = 0; m < M; ++m)
for (size_t n = 0; n < N; ++n)
Destination[m*N+n] = Source[n*M+m];
}
The above requires the caller to cast the arguments to float *. We can make it more convenient for the caller with:
void matrix_transposition(size_t M, size_t N,
void *DestinationPointer, const void *SourcePointer)
{
float *Destination = DestinationPointer;
const float *Source = SourcePointer;
for (size_t m = 0; m < M; ++m)
for (size_t n = 0; n < N; ++n)
Destination[m*N+n] = Source[n*M+m];
}
(Unfortunately, this prevents the compiler from checking that the argument types match the intended types, but this is a shortcoming of C.)
If you need a solution strictly in standard C without variable length arrays, then, technically, the proper way is to copy the bytes of the objects:
void matrix_transposition(size_t M, size_t N,
void *DestinationPointer, const void *SourcePointer)
{
char *Destination = DestinationPointer;
const char *Source = SourcePointer;
for (size_t m = 0; m < M; ++m)
for (size_t n = 0; n < N; ++n)
{
// Calculate locations of elements in memory.
char *D = Destination + (m*N+n) * sizeof(float);
const char *S = Source + (n*M+m) * sizeof(float);
memcpy(D, S, sizeof(float));
}
}
Notes:
Include <stdlib.h> to declare size_t and, if using the last solution, include <string.h> to declare memcpy.
Variable length arrays were required in C 1999 but made optional in C 2011. Good quality compilers for general purpose systems will support them.
If you are using C99 compiler, you can make use of Variable Length Array (VLA's) (optional in C11 compiler). You can write a function like this:
void matrix_transposition (int rows_in_matrix, int columnes_in_matrix, float transposed_matrix[columnes_in_matrix][rows_in_matrix], float matrix[rows_in_matrix][columnes_in_matrix])
{
int i,j;
for(i = 0; i < rows_in_matrix; ++i) {
for(j = 0; j < columnes_in_matrix; ++j)
{
transposed_matrix[j][i] = matrix[i][j];
}
}
}
This one function can work for the different number of rows_in_matrix and columnes_in_matrix. Call it like this:
matrix_transposition (2, 4, transposed_matrix_A, input_matrix_A);
matrix_transposition (4, 4, transposed_matrix_B, input_matrix_B);
You probably don't want to be hard-coding array sizes in your program. I suggest a structure that contains a single flat array, which you can then interpret in two dimensions:
typedef struct {
size_t width;
size_t height;
float *elements;
} Matrix;
Initialize it with
int matrix_init(Matrix *m, size_t w, size_t h)
{
m.elements = malloc((sizeof *m.elements) * w * h);
if (!m.elements) {
m.width = m.height = 0;
return 0; /* failed */
}
m.width = w;
m.height = h;
return 1; /* success */
}
Then, to find the element at position (x,y), we can use a simple function:
float *matrix_element(Matrix *m, size_t x, size_t y)
{
/* optional: range checking here */
return m.elements + x + m.width * y;
}
This has better locality than an array of pointers (and is easier and faster to allocate and deallocate correctly), and is more flexible than an array of arrays (where, as you've found, the inner arrays need a compile-time constant size).
You might be able to use an array of arrays wrapped in a Matrix struct - it's possible you'll need a stride that is not necessarily the same as width, if the array of arrays has padding on your platform.
I am trying to make a grid of blank spaces with 'x' dotted around randomly. I have managed to do this but now I want a separate function to print out the 2d array but am finding it hard to do so... any help will be great, thanks in advance.
int* createEnvironment(int width, int height, int numOfX)
{
int i, j;
char array[width][height];
for (i = 0; i < width; i++)
{
for( j = 0; j < height; j++)
{
array[i][j] = '.';
}
}
srand(time(NULL));
for(i=0; i<numOfX; i++)
{
int row = ((rand() % (width-1)) + 0);
int col = ((rand() % (height-1)) + 0);
array[row][col] = 'x';
}
return array;
}
then print out 2d array...
void printEnvironment(int* environment, int width, int height)
{
int i, j;
for (i = 0; i < (height-1) ; i++)
{
printf("\n");
for( j = 0; j < (width-1); j++)
{
printf("%c", environment[i][j]);
}
}
}
int main(int argc, char *argv[])
{
int height = 20;
int width = 80;
int numOfX = 100;
int* environment;
environment = createEnvironment(width, height , numOfX);
printEnvironment(environment, width, height);
system("PAUSE");
return 0;
}
(It looks like you have a stray }, but that's not relevant to the answer)
In your function int* createEnvironment(int width, int height, int numOfX), when you do char array[width][height];, you allocate a width x height block of memory in the stack. Your then use a few for loops that fill in the X's in certain spots. That all looks fine.
The problem is that array is allocated on the stack. Once createEnvironment returns, all of the variables that were placed on the stack will no longer be valid, since those areas of memory might be used by subsequent function calls. An easier way to think about this is to consider where each variable 'lives' within the program. Global variables live in the program and can be accessed by anyone. Variables that live in main can be accessed in main, and in functions that are called by main (i.e. with pointers). However, if you call function A from main, which has some variable V declared, and then call function B from main, B cannot access V, since V only lived in main. The solution to this problem would be to pass V back to main to be sent to B.
Fortunately, the C standard library provides malloc, which allocates memory off the heap, not the stack. When you allocate memory in the heap, it live in the heap regardless of which context it is created from. If function A creates variable V in the heap with malloc, then V can be accessed from main, B, or any other function, so long as a pointer to V is provided. This is also where memory leaks come from; when space is allocated from the heap, if it is not freed with free, then it stays allocated until the program ends.
The short answer is to replace the line char array[width][height]; with char * array = malloc(width * height * sizeof(char));. Not that the sizeof isn't strictly necessary since a char is one byte in size, but that would need to be changed if you changed char to some other type.
You'll also need to update the way you access a sub-element of the array. Since the width and height of the array won't be known if it's in the heap, you'll have to instead provide an offset from array, which should give you a clue how arrays work in C to begin with. I'll leave that for you to figure out.
I would like to store values read from a for-loop to an array
char A[];
int x;
int y=5;
for( int i=0; int i =1000; i++) {
x = x+y;
// then store/append x as elements of the char array, A.... what is the syntax?
}
By looking at your code I am assuming that you are trying to build a static array, so I will demonstrate that (so you don't have to focus on concepts like malloc for the time being). There is however, several problems with your code that I will go over now.
First off your array declaration:
char A[];
to me it looks like your for loop is filling an array of integers, so this array should be declared as an integer, furthermore you are not setting the size of the array, since your code has i increment until it is 1000 you should just declare an integer array with 1000 elements:
int A[1000];
Second your for loop:
for(int i = 0, int i = 1000; i++)
you're better off just declaring i with the rest of your variables, although you can declare it in a for loop I personally wouldn't suggest doing it. Also you declare i twice in this loop. Finally your condition to continue the loop (i = 1000) will abort the loop immediatly since i will never be equal to 1000 since you set it to 0. Remember a for loop only loops while the middle statement is true. So with that in mind you should now have:
int A[1000], i, x, y = 5;
for(i = 0; i < 1000; i++)
And now we can use the = statement and the value of i to set each array element for A:
int A[1000], i, x, y = 5;
for(i = 0; i < 1000; i++)
{
x += y;
A[i] = x;
}
it's that simple!
There are multiple issues with your code
char A[1000]; // Need to specify a compile time constant for the array size
int x=0;
int y=5;
for( int i=0; i < 1000; i++) { // Condition was wrong
x = x+y;
// then store/append x as elements of the char array, A.... what is the syntax?
A[i] = x; // Add the value
}
Also, the char datatype won't be able to hold values over a certain size, and will cause overflow making the values wrap around. You might want to declare A as int A[1000] instead.
Arrays need to be of a constant size, or you will need to allocate them using malloc
The second part of the loop cannot redeclare i again. It also will loop forever if you have an assignment statement in it like you do. I assume you want to loop up to 1000 instead
The actual question, to assign into an array use the [] operator.
x was not initialized to anything, making it contain a garbage value. You need to assign values to variables upon declaring them. C does not do this for you automatically.
If you want to add an element in C, you have several methods.
Static array
A static array is declared with a number of elements you're unable to edit. So it's perfect if you know exactly the number of elements you'll have. #Dervall did explain that well.
Dynamic array
A dynamic array is declared with malloc function. And the size can be changed. It's difficult and hard to maintain though. But :
int *A = NULL;
int *tmp; // to free ex allocated arrays
int i;
int j;
int x = 0;
int y = 5;
for (i = 0 ; i < 1000 ; i++) {
// saving temporarly the ex array
tmp = A;
// we allocate a new array
if ((A = malloc(sizeof(int) * (i + 1))) == NULL) {
return EXIT_FAILURE;
}
// we fill the new array allocated with ex values which are in tmp
for (j = 0; j < i; j++) {
A[j] = tmp[j];
}
// if it's not the first time, we free the ex array
if (tmp != NULL)
free(tmp);
x = x + y;
A[i] = x;
}
Better to split it into a function of course :)
You can use the realloc function as well ! Which is made for that, but I find it interesting to develop like this
There's a lot of stuff wrong with your snippet. Here's a compilable example
char *A = malloc(sizeof(*A) * NUM_ELEMENTS); // you shouldn't declare on the stack
int x = 0; // initialize
int y=5;
for( int i = 0; i < NUM_ELEMENTS; i++) { // proper for loop syntax
x = x+y;
A[i]=x; // assign element of array
}
And a better version:
char *A = malloc(sizeof(*A) * NUM_ELEMENTS);
for (int i = 0; i < NUM_ELEMENTS; ++i)
A[i] = 5 * i;
I want to create a program in which I can pass a matrix to a function using pointers.
I initialized and scanned 2 matrices in the void main() and then I tried to pass them to a void add function. I think I am going wrong in the syntax of declaration and calling of the function. I assigned a pointer to the base address of my matrix. (for eg: int *x=a[0][0], *y=b[0][0]). What is the right declaration? How can I specify the dimensions?
Given a 2D array of
T a[N][M];
a pointer to that array would look like
T (*ap)[M];
so your add function prototype should look like
void add(int (*a)[COLS], int (*b)[COLS]) {...}
and be called as
int main(void)
{
int a[ROWS][COLS];
int b[ROWS][COLS];
...
add(a, b);
However, this code highlights several problems. First is that your add function is relying on information not passed via the parameter list, but via a global variable or symbolic constant; namely, the number of rows (the number of columns is explicitly provided in the type of the parameters). This tightly couples the add function to this specific program, and makes it hard to reuse elsewhere. For your purposes this may not be a problem, but in general you only want your functions to communicate with their callers through the parameter list and return values.
The second problem is that as written, your function will only work for matrices of ROWS rows and COLS columns; if you want to add matrices of different sizes within the same program, this approach will not work. Ideally you want an add function that can deal with matrices of different sizes, meaning you need to pass the sizes in as separate parameters. It also means we must change the type of the pointer that we pass in.
One possible solution is to treat your matrices as simple pointers to int and manually compute the offsets instead of using subscripts:
void add (int *a, int *b, size_t rows, size_t cols)
{
size_t i;
for (i = 0; i < rows; i++)
{
size_t j;
for (j = 0; j < cols; j++)
{
*(a + cols * i + j) += *(b + cols * i + j);
}
}
}
and call it like so:
int main(void)
{
int a[ROWS][COLS] = {...};
int b[ROWS][COLS] = {...};
int c[ROWS2][COLS2] = {...};
int d[ROWS2][COLS2] = {...};
...
add(a[0], b[0], ROWS, COLS);
add(c[0], d[0], ROWS2, COLS2);
...
}
The types of a[0] and b[0] are "COLS-element arrays of int"; in this context, they'll both be implicitly converted to "pointer to int". Similarly, c[0] and d[0] are also implicitly converted to int *. The offsets in the add() function work because 2D arrays are contiguous.
EDIT I just realized I was responding to caf's example, not the OP, and caf edited his response to show something very similar to my example. C'est la guerre. I'll leave my example as is just to show a slightly different approach. I also think the verbiage about passing information between functions and callers is valuable.
Something like this should do the trick.
#define COLS 3
#define ROWS 2
/* Store sum of matrix a and b in a */
void add(int a[][COLS], int b[][COLS])
{
int i, j;
for (i = 0; i < ROWS; i++)
for (j = 0; j < COLS; j++)
a[i][j] += b[i][j];
}
int main()
{
int a[ROWS][COLS] = { { 5, 10, 5} , { 6, 4, 2 } };
int b[ROWS][COLS] = { { 2, 3, 4} , { 10, 11, 12 } };
add(a, b);
return 0;
}
EDIT: Unless you want to specify the dimensions at runtime, in which case you have to use a flat array and do the 2D array arithmetic yourself:
/* Store sum of matrix a and b in a */
void add(int rows, int cols, int a[], int b[])
{
int i, j;
for (i = 0; i < rows; i++)
for (j = 0; j < cols; j++)
a[i * cols + j] += b[i * cols + j];
}
#caf has shown a good code example.
I'd like to point out that:
I assigned a pointer to the base
address of my matrix. (for eg: int
*x=a[0][0],*y=b[0][0]).
You are not assining a pointer to the base of the matrix. What this does is assign to the value pointed by x and y, the base value in a and b respectively.
The right way would be
int (*x)[] = a;
int (*y)[] = b;
or alternatively
int *x = &a[0][0];
int *y = &b[0][0];