Differences between matrix implementation in C - c

I created two 2D arrays (matrix) in C in two different ways.
I don't understand the difference between the way they're represented in the memory, and the reason why I can't refer to them in the same way:
scanf("%d", &intMatrix1[i][j]); //can't refer as &intMatrix1[(i * lines)+j])
scanf("%d", &intMatrix2[(i * lines)+j]); //can't refer as &intMatrix2[i][j])
What is the difference between the ways these two arrays are implemented and why do I have to refer to them differently?
How do I refer to an element in each of the arrays in the same way (?????? in my printMatrix function)?
int main()
{
int **intMatrix1;
int *intMatrix2;
int i, j, lines, columns;
lines = 3;
columns = 2;
/************************* intMatrix1 ****************************/
intMatrix1 = (int **)malloc(lines * sizeof(int *));
for (i = 0; i < lines; ++i)
intMatrix1[i] = (int *)malloc(columns * sizeof(int));
for (i = 0; i < lines; ++i)
{
for (j = 0; j < columns; ++j)
{
printf("Type a number for intMatrix1[%d][%d]\t", i, j);
scanf("%d", &intMatrix1[i][j]);
}
}
/************************* intMatrix2 ****************************/
intMatrix2 = (int *)malloc(lines * columns * sizeof(int));
for (i = 0; i < lines; ++i)
{
for (j = 0; j < columns; ++j)
{
printf("Type a number for intMatrix2[%d][%d]\t", i, j);
scanf("%d", &intMatrix2[(i * lines)+j]);
}
}
/************** printing intMatrix1 & intMatrix2 ****************/
printf("intMatrix1:\n\n");
printMatrix(*intMatrix1, lines, columns);
printf("intMatrix2:\n\n");
printMatrix(intMatrix2, lines, columns);
}
/************************* printMatrix ****************************/
void printMatrix(int *ptArray, int h, int w)
{
int i, j;
printf("Printing matrix...\n\n\n");
for (i = 0; i < h; ++i)
for (j = 0; j < w; ++j)
printf("array[%d][%d] ==============> %d\n, i, j, ??????);
}

You are dereferencing the Matrix1 two times..
Matrix1[i][j] ;
It means that it is a 2D array or a double pointer declared like this.
int **Matrix1 ;
A double pointer could be thought of as array of pointers. Its each element is a pointer itself, so it is dereferenced once to reach at the pointer element, and dereferenced twice to access the data member of that member pointer or array. This statement as you you wrote is equivalent to this one..
Matrix1[i][j] ; //is ~ to
*( *(Matrix1 + i) + j) ;
For a single pointer like this.
int *Matrix2 ;
You can derefernce it only once, like this.
Matrix2[i] ; //is ~ to
*(Matrix2 + i) ;
This statement which you wrote..
Matrix2[(i * lines)+j] ;
|-----------|
This portion evaluates to a single number, so it derefenced one time.
(i * lines) + j ;
As for your printmatrix() function, the ptArray passed to it is a single pointer. So you cannot dereference it twice.
Perhaps you can get better understanding of static and dynamic 2D arrays from my answer here.
2D-array as argument to function

Both matrices are sequences of bytes in memory. However, the difference between them is how you're defining the memory interface to represent a matrix. In one case you're just defining a memory segment with a number of elements equal to the elements in the matrix, and in the other case you're specifically allocating memory to represent each specific line.
The following case is more expensive computationally, because you're invoking malloc() a greater number of times:
intMatrix1 = (int **)malloc(lines * sizeof(int *));
for (i = 0; i < lines; ++i)
intMatrix1[i] = (int *)malloc(columns * sizeof(int));
However, it brings the advantage that you get to refer to matrix elements in a clearer fashion:
intMatrix1[i][j];
If you just allocate one sequence of elements equal to the number of elements in the matrix, you have to take in account line/column index calculations to refer to the right matrix elements in memory.
To attempt to increase the degree of uniformity in the code, may I suggest a function that receives the matrix line reference and matrix column-count and prints a line?
void PrintLine(int *ptrLine, int lineLen) {
unsigned int i;
for(i = 0; i < lineLen; i++)
printf("%d ", ptrLine[i]);
printf("\n");
}
And then, for each matrix type, you would just do:
// Case 1
for(i = 0; i < lines; i++)
PrintLine(intMatrix1[i], columns);
// Case 2
for(i = 0; i < lines; i++) {
PrintLine(intMatrix2 + i*columns, columns);
}

In C, the array access operator [] is really just a cleaner way of performing pointer arithmetic. For a one-dimensional array of elements of type type_s, arr[i] is equivalent to *(arr + (i * sizeof(type_s))). To dissect that expression:
arr will be the base address, the lowest memory address where this array is stored
i is the zero-indexed position of the element in the array
sizeof returns the number of chars (which is generally the same as the number of bytes, but it's not mandated by the C spec) that an element in arr takes up in memory. The compiler will determine the size of the element and take care of performing this math for you.
As a side note, this syntax has the side effect of arr[i] being equivalent to i[arr], although it's universally accepted to put the index in brackets.
So with all of that said, let's look at the differences between your two declarations:
intMatrix1[i][j] is equivalent to *(*(intMatrix1 + i * sizeof(int)) + j * sizeof(int)). So, there are two dereference operators in that expression, meaning that intMatrix is an array of arrays (it contains pointers to pointers).
On the other hand, intMatrix2[(i * lines)+j] is equivalent to *(intMatrix2 + ((i * lines) + j) * sizeof(int)), which contains only one dereference operator. What you're doing here is defining a one-dimensional array that contains the same number of elements as the original two-dimensional array. If your data can be best represented by a matrix, then I recommend you use the first version: intMatrix1[i][j].

The difference is that the first array:
intMatrix1 = (int **)malloc(lines * sizeof(int *));
Creates an array of pointers intMatrix1. Each of those pointers points to an int array (which you malloc here).
for (i = 0; i < lines; ++i)
intMatrix1[i] = (int *)malloc(columns * sizeof(int));
That's why you need the 2 stars (dereference to the pointer array, then to the int array) in the declaration and the double brackets to access single elements:
int **intMatrix1;
int i = intMatrix[row][column];
int i = *(*(intmatrix + row) + column);
For the second matrix, you create just an int array of size column * rows.
int *intMatrix2 = (int *)malloc(lines * columns * sizeof(int));
int i = intMatrix[row + column];
int i = *(intMatrix + row + column);
To print the 2 arrays you will have to use different print functions, because the internal structure of the 2 matrix is different, but you already know the different methods to access both arrays.

Related

How many pointers are in an array of pointers

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)

C Pointer to Pointer Problems in Matrix Function

I know there are very similar questions, but I've read them and I don't understand what is going wrong or what exactly I haven't understood about pointers to pointers.
My teacher is using a "learning by doing" approach to pointers and it has left me with approximately 100 questions. Sometimes I just change things around until it compiles, but it really isn't becoming any more clear to me, so if someone could help clarify a few things, I'd really appreciate it.
struct Matrix {
int rows; // number of rows
int cols; // number of columns
double **data;
};
typedef struct Matrix Matrix;
The pointer to a pointer, is something like this, right?
double *row1 = (double *) malloc (n_cols*sizeof(double));
double *row2 = (double *) malloc (n_cols*sizeof(double));
double *row3 = (double *) malloc (n_cols*sizeof(double));
double *data[] = { row1, row2, row3};
Data is pointing to the row number which is pointing to the doubles in the rows.
Now I am supposed to make a constructor function that makes a Matrix with a 0 in every position returns a pointer to a Matrix.
Matrix *make_matrix(int n_rows, int n_cols) {
Matrix *m = xmalloc(sizeof(Matrix));
m->rows = n_rows;
m->cols = n_cols;
double **rows_and_columns = xmalloc(n_rows * n_cols * sizeof(double));
memset(rows_and_columns, 0, m->rows * m->cols * sizeof(double));
m->data = *rows_and_columns;
return m;
}
So I made a pointer for the matrix, then I assigned the values for the rows and columns. Then I got confused (although I am not sure how confused, because this part compiles). I made a pointer to pointer for the last element of the struct Matrix (**data). Since **rows_and_columns has to hold the rows and columns, I allocated xmalloc(n_rows * n_cols * sizeof(double)) memory to it. I then set the whole thing to 0 and assign it to data. I think this m->data = rows_and_columns; says that m points to data and since data is a pointer and rows_and_columns is a pointer, we'll align their addresses, so m->data will also point to a bunch of 0s? Or is this wrong? And I am returning m, because the output is Matrix * and the m will get the * upon output, correct?
The next step is to copy a matrix, at which point I got even more lost.
Matrix *copy_matrix(double *data, int n_rows, int n_cols) {
Matrix *m = make_matrix(n_rows, n_cols);
double *row = (double *) malloc (n_cols*sizeof(double));
int i = 0;
for (int j = 0; j < n_rows; j++) {
for (; i < n_cols; i++) {
row = (double *) malloc (n_cols*sizeof(double));
row [i] = data[i + j*n_cols];
}
m->data [j] = *row [i];
}
free(row);
return m;
}
So we are returning a pointer to a Matrix again. The input is now a pointer to double values, which I am assuming are the rows. First, I made a matrix. Then I made a pointer for a row with a n columns worth of memory (double *) malloc (n_cols*sizeof(double)).
This is where I got super confused. I was imagining **data the whole time as something like above (double *data[] = { row1, row2, row3};). So, I wanted to copy each row of *data into *row, then save that as an entry in **data, like data[0] = row0, but something isn't clicking with the pointers, because I am not allowed to assign m->data [j] = row [i];, because I'm assigning incompatible types by assigning double * from type double?
xmalloc() returns a void * pointer to single block of memory.
What you need is one block of pointers, serving as an conceptual table header row, holding pointers to other memory blocks which themselves contain the actual doubles.
double **columns -> [double *col0] [double *col1] [double *col2] ...
| | |
V V V
[double col_val0] [double col_val0] ...
[double col_val1] [double col_val1]
[double col_val2] [double col_val2]
... ...
A matrix allocation could look like this:
// Allocate the double pointer array:
double **matrix_rows = xmalloc(sizeof(double *) * column_count);
// Iterate over each double-pointer in the double-pointer-array allocated above.
for(int i = 0; i < column_count; i++) {
// Allocate a new double-array, and let current double-pointer point to it:
matrix_rows[i] = malloc(sizeof(double) * row_count);
// Initialize matrix cell, iterating over allocated values.
for(int j = 0; j < row_count; j++) {
// Accessing the i'th col and j'th cell.
matrix_rows[i][j] = 0.0;
}
}
A possible implementation of a matrix copy function could be done by iteratively copying individial cells. One way to do this is using a loop composition.
for(int col = 0; col < col_count; col++) {
for(int row = 0; row < row_count; row++) {
destination_matrix[col][row] = source_matrix[col][row];
}
}
To give some intuition where an n-pointer indirection could be used:
n = 1: Strings, an array of characters.
n = 2: Paragraph, holding lines of strings.
n = 3: An article, holding a list of paragraphs.
Please be aware of using two indirections is usually not something you want. It is usually more efficient to store data in a linear fashion and compute linear indices out of two-compoment vectors and the other way around, especially in the case of this matrix example.
If you want to represent a matrix as an array of pointers to rows you need to allocate memory both for the rows and for the array of pointers to rows. It is simpler to allocate the rows consecutively as a single block.
typedef struct
{
int n_rows;
int n_cols;
double **rows;
double *data;
} Matrix;
Matrix *matrix_new (int n_rows, int n_cols)
{
// allocate matrix
Matrix *m = malloc (sizeof(Matrix));
m->n_rows = n_rows;
m->n_cols = n_cols;
// allocate m->data
m->data = malloc (n_rows * n_cols * sizeof(double));
// allocate and fill m->rows
m->rows = malloc (n_rows * sizeof(double*));
for (int i = 0; i < n_rows; i++) {
m->rows[i] = &data[i * n_cols];
}
// set the matrix to 0
for (int i = 0; i < n_rows; i++) {
for (int j = 0; j < n_cols; j++) {
m->rows[i][j] = 0.0;
}
}
return m;
}
The purpose of the rows array it to give you the convenience of being able to refer to element i,j with m->rows[i][j] instead of m->data[i * m->n_cols + j].
To free the matrix, take the inverse steps:
void matrix_free (Matrix *m)
{
free (m->rows);
free (m->data);
free (m);
}
To copy you can simply allocate a matrix of the same size and copy element by element:
Matrix *matrix_copy (Matrix *m1)
{
Matrix *m2 = matrix_new (m1->n_rows, m1->n_cols);
for (int i = 0; i < m1->n_rows; i++) {
for (int j = 0; j < m1->n_cols; j++) {
m2->rows[i][j] = m1->rows[i][j];
}
}
return m2;
}
The important thing to note is that you must not copy the rows array since it is unique to each matrix.
It is important to understand the difference between pointers-to-pointers and multi-dimensional arrays.
What makes it extra confusing is that the same syntax is used for referencing individual elements: var[i][j] will reference element (i,j) of var regardless of if var is a pointer to pointer, double **var or a two-dimensional array, double var[22][43].
What really happens is not the same. A two-dimensional array is a contiguous memory block. A pointer to pointers is an array of pointers that point to the individual rows.
// Two-dimensional array
char var1[X1][X2];
int distance = &var[4][7] - &var[0][0]; // distance = 4*X2+7
// Pointer-to-pointer
char **var2 = malloc(X1 * sizeof(char*)); // Allocate memory for the pointers
for (i=0; i<X1; i++) var2[i] = malloc(X2); // Allocate memory for the actual data
int distance2 = &var2[4][7] - &var[0][0]; // This could result in anything, since the different rows aren't necessarily stored one after another.
The calculation of distance2 invokes undefined behaviour since the C standard doesn't cover pointer arithmetic on pointers that point to different memory blocks.
You want to use pointer-to-pointer. So you need to first allocate memory for an array of pointers and then for each array:
Matrix *make_matrix(int n_rows, int n_cols) {
Matrix *m = xmalloc(sizeof(Matrix));
int i, j;
m->rows = n_rows;
m->cols = n_cols;
m->data = xmalloc(n_rows * sizeof(double*));
for (i=0; i < n_; i++) {
m->data[i] = xmalloc(n_cols * sizeof(double));
for (j=0; j < n_cols; j++) {
m->data[i][j] = 0.0;
}
}
return m;
}
Don't assume that the double 0.0 will have all bits set to 0!
To copy a matrix:
Matrix *copy_matrix(Matrix *source) {
Matrix *m = make_matrix(source->n_rows, source->n_cols);
int i, j;
for (j = 0; j < n_rows; j++) {
for (i = 0; i < n_cols; i++) {
m->data[i][j] = source[i][j];
}
}
return m;
}
I'll backup a bit and start with the basics. Pointers are one of those things that are not difficult to understand technically, but require you to beat your head into the I want to understand pointers wall enough for them to sink in. You understand that a normal variable (for lack of better words) is just a variable that holds a direct-reference to an immediate value in memory.
int a = 5;
Here, a is a label to the memory address that holds the value 5.
A pointer on the other hand, does not directly-reference an immediate value like 5. Instead a pointer holds, as its value, the memory address where 5 is stored. You can also think of the difference this way. A normal variable holds a value, while a pointer holds the address where the value can be found.
For example, to declare a pointer 'b' to the memory address holding 5 above, you would do something like:
int *b = &a;
or equivalently:
int *b = NULL;
b = &a;
Where b is assigned the address of a. To return the value stored at the address held by a pointer, or to operate directly on the value stored at the address held by a pointer, you must dereference the pointer. e.g.
int c = *b; /* c now equals 5 */
*b = 7; /* a - the value at the address pointed to by b, now equals 7 */
Now fast forward to pointer-to-pointer-to-type and simulated 2D matricies. When you declare your pointer-to-pointer-to-type (e.g. int **array = NULL), you are declaring a pointer that points to a pointer-to-type. To be useful in simlated 2D arrays (matricies, etc.), you must delcare an array of the pointer-to-pointer-to-type:
int **array = NULL;
...
array = calloc (NUM, sizeof *array); /* creates 'NUM' pointer-to-pointer-to-int. */
You now have NUM pointers to pointers-to-int that you can allocate memory to hold however many int values and you will assign the starting address for the memory block holding those int values to the pointers you previously allocated. For example, let's say you were to allocate space for an array of 5 random int values (from 1 - 1000) to each of the NUM pointers you allocated above:
for (i = 0; i < NUM; i++) {
array[i] = calloc (5, sizeof **array);
for (j = 0; j < 5; j++)
array[i][j] = rand() % 1000 + 1;
}
You have now assigned each of your NUM pointers (to-pointer-to-int) the starting address in memory where 5 random int values are stored. So your array is now complete. Each of your original NUM pointers-to-pointer-to-int now points to the address for a block of memory holding 5 int values. You can access each value with array notation (e.g. array[i][j] or with pointer notation *(*(array + i) + j) )
How do you free the memory? In the reverse order you allocated (values, then pointers):
for (i = 0; i < NUM; i++)
free (array[i]);
free (array);
Note: calloc both allocates memory and initializes the memory to 0/nul. This is particularly useful for both the pointers and arrays when dealing with numeric values, and when dealing with an unknown number of rows of values to read. Not only does it prevent an inadvertent read from an uninitialized value, but it also allows you to iterate over your array of pointers with i = 0; while (array[i] != NULL) {...}.

Dynamically creating a 2 dimensional array of pointers in c [duplicate]

Does someone know how I can use dynamically allocated multi-dimensional arrays using C? Is that possible?
Since C99, C has 2D arrays with dynamical bounds. If you want to avoid that such beast are allocated on the stack (which you should), you can allocate them easily in one go as the following
double (*A)[n] = malloc(sizeof(double[n][n]));
and that's it. You can then easily use it as you are used for 2D arrays with something like A[i][j]. And don't forget that one at the end
free(A);
Randy Meyers wrote series of articles explaining variable length arrays (VLAs).
With dynamic allocation, using malloc:
int** x;
x = malloc(dimension1_max * sizeof(*x));
for (int i = 0; i < dimension1_max; i++) {
x[i] = malloc(dimension2_max * sizeof(x[0]));
}
//Writing values
x[0..(dimension1_max-1)][0..(dimension2_max-1)] = Value;
[...]
for (int i = 0; i < dimension1_max; i++) {
free(x[i]);
}
free(x);
This allocates an 2D array of size dimension1_max * dimension2_max. So, for example, if you want a 640*480 array (f.e. pixels of an image), use dimension1_max = 640, dimension2_max = 480. You can then access the array using x[d1][d2] where d1 = 0..639, d2 = 0..479.
But a search on SO or Google also reveals other possibilities, for example in this SO question
Note that your array won't allocate a contiguous region of memory (640*480 bytes) in that case which could give problems with functions that assume this. So to get the array satisfy the condition, replace the malloc block above with this:
int** x;
int* temp;
x = malloc(dimension1_max * sizeof(*x));
temp = malloc(dimension1_max * dimension2_max * sizeof(x[0]));
for (int i = 0; i < dimension1_max; i++) {
x[i] = temp + (i * dimension2_max);
}
[...]
free(temp);
free(x);
Basics
Arrays in c are declared and accessed using the [] operator. So that
int ary1[5];
declares an array of 5 integers. Elements are numbered from zero so ary1[0] is the first element, and ary1[4] is the last element. Note1: There is no default initialization, so the memory occupied by the array may initially contain anything. Note2: ary1[5] accesses memory in an undefined state (which may not even be accessible to you), so don't do it!
Multi-dimensional arrays are implemented as an array of arrays (of arrays (of ... ) ). So
float ary2[3][5];
declares an array of 3 one-dimensional arrays of 5 floating point numbers each. Now ary2[0][0] is the first element of the first array, ary2[0][4] is the last element of the first array, and ary2[2][4] is the last element of the last array. The '89 standard requires this data to be contiguous (sec. A8.6.2 on page 216 of my K&R 2nd. ed.) but seems to be agnostic on padding.
Trying to go dynamic in more than one dimension
If you don't know the size of the array at compile time, you'll want to dynamically allocate the array. It is tempting to try
double *buf3;
buf3 = malloc(3*5*sizeof(double));
/* error checking goes here */
which should work if the compiler does not pad the allocation (stick extra space between the one-dimensional arrays). It might be safer to go with:
double *buf4;
buf4 = malloc(sizeof(double[3][5]));
/* error checking */
but either way the trick comes at dereferencing time. You can't write buf[i][j] because buf has the wrong type. Nor can you use
double **hdl4 = (double**)buf;
hdl4[2][3] = 0; /* Wrong! */
because the compiler expects hdl4 to be the address of an address of a double. Nor can you use double incomplete_ary4[][]; because this is an error;
So what can you do?
Do the row and column arithmetic yourself
Allocate and do the work in a function
Use an array of pointers (the mechanism qrdl is talking about)
Do the math yourself
Simply compute memory offset to each element like this:
for (i=0; i<3; ++i){
for(j=0; j<3; ++j){
buf3[i * 5 + j] = someValue(i,j); /* Don't need to worry about
padding in this case */
}
}
Allocate and do the work in a function
Define a function that takes the needed size as an argument and proceed as normal
void dary(int x, int y){
double ary4[x][y];
ary4[2][3] = 5;
}
Of course, in this case ary4 is a local variable and you can not return it: all the work with the array must be done in the function you call of in functions that it calls.
An array of pointers
Consider this:
double **hdl5 = malloc(3*sizeof(double*));
/* Error checking */
for (i=0; i<3; ++i){
hdl5[i] = malloc(5*sizeof(double))
/* Error checking */
}
Now hdl5 points to an array of pointers each of which points to an array of doubles. The cool bit is that you can use the two-dimensional array notation to access this structure---hdl5[0][2] gets the middle element of the first row---but this is none-the-less a different kind of object than a two-dimensional array declared by double ary[3][5];.
This structure is more flexible then a two dimensional array (because the rows need not be the same length), but accessing it will generally be slower and it requires more memory (you need a place to hold the intermediate pointers).
Note that since I haven't setup any guards you'll have to keep track of the size of all the arrays yourself.
Arithmetic
c provides no support for vector, matrix or tensor math, you'll have to implement it yourself, or bring in a library.
Multiplication by a scaler and addition and subtraction of arrays of the same rank are easy: just loop over the elements and perform the operation as you go. Inner products are similarly straight forward.
Outer products mean more loops.
If you know the number of columns at compile time, it's pretty simple:
#define COLS ...
...
size_t rows;
// get number of rows
T (*ap)[COLS] = malloc(sizeof *ap * rows); // ap is a *pointer to an array* of T
You can treat ap like any 2D array:
ap[i][j] = x;
When you're done you deallocate it as
free(ap);
If you don't know the number of columns at compile time, but you're working with a C99 compiler or a C2011 compiler that supports variable-length arrays, it's still pretty simple:
size_t rows;
size_t cols;
// get rows and cols
T (*ap)[cols] = malloc(sizeof *ap * rows);
...
ap[i][j] = x;
...
free(ap);
If you don't know the number of columns at compile time and you're working with a version of C that doesn't support variable-length arrays, then you'll need to do something different. If you need all of the elements to be allocated in a contiguous chunk (like a regular array), then you can allocate the memory as a 1D array, and compute a 1D offset:
size_t rows, cols;
// get rows and columns
T *ap = malloc(sizeof *ap * rows * cols);
...
ap[i * rows + j] = x;
...
free(ap);
If you don't need the memory to be contiguous, you can follow a two-step allocation method:
size_t rows, cols;
// get rows and cols
T **ap = malloc(sizeof *ap * rows);
if (ap)
{
size_t i = 0;
for (i = 0; i < cols; i++)
{
ap[i] = malloc(sizeof *ap[i] * cols);
}
}
ap[i][j] = x;
Since allocation was a two-step process, deallocation also needs to be a two-step process:
for (i = 0; i < cols; i++)
free(ap[i]);
free(ap);
malloc will do.
int rows = 20;
int cols = 20;
int *array;
array = malloc(rows * cols * sizeof(int));
Refer the below article for help:-
http://courses.cs.vt.edu/~cs2704/spring00/mcquain/Notes/4up/Managing2DArrays.pdf
Here is working code that defines a subroutine make_3d_array to allocate a multidimensional 3D array with N1, N2 and N3 elements in each dimension, and then populates it with random numbers. You can use the notation A[i][j][k] to access its elements.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
// Method to allocate a 2D array of floats
float*** make_3d_array(int nx, int ny, int nz) {
float*** arr;
int i,j;
arr = (float ***) malloc(nx*sizeof(float**));
for (i = 0; i < nx; i++) {
arr[i] = (float **) malloc(ny*sizeof(float*));
for(j = 0; j < ny; j++) {
arr[i][j] = (float *) malloc(nz * sizeof(float));
}
}
return arr;
}
int main(int argc, char *argv[])
{
int i, j, k;
size_t N1=10,N2=20,N3=5;
// allocates 3D array
float ***ran = make_3d_array(N1, N2, N3);
// initialize pseudo-random number generator
srand(time(NULL));
// populates the array with random numbers
for (i = 0; i < N1; i++){
for (j=0; j<N2; j++) {
for (k=0; k<N3; k++) {
ran[i][j][k] = ((float)rand()/(float)(RAND_MAX));
}
}
}
// prints values
for (i=0; i<N1; i++) {
for (j=0; j<N2; j++) {
for (k=0; k<N3; k++) {
printf("A[%d][%d][%d] = %f \n", i,j,k,ran[i][j][k]);
}
}
}
free(ran);
}
There's no way to allocate the whole thing in one go. Instead, create an array of pointers, then, for each pointer, create the memory for it. For example:
int** array;
array = (int**)malloc(sizeof(int*) * 50);
for(int i = 0; i < 50; i++)
array[i] = (int*)malloc(sizeof(int) * 50);
Of course, you can also declare the array as int* array[50] and skip the first malloc, but the second set is needed in order to dynamically allocate the required storage.
It is possible to hack a way to allocate it in a single step, but it would require a custom lookup function, but writing that in such a way that it will always work can be annoying. An example could be L(arr,x,y,max_x) arr[(y)*(max_x) + (x)], then malloc a block of 50*50 ints or whatever and access using that L macro, e.g.
#define L(arr,x,y,max_x) arr[(y)*(max_x) + (x)]
int dim_x = 50;
int dim_y = 50;
int* array = malloc(dim_x*dim_y*sizeof(int));
int foo = L(array, 4, 6, dim_x);
But that's much nastier unless you know the effects of what you're doing with the preprocessor macro.
int rows, columns;
/* initialize rows and columns to the desired value */
arr = (int**)malloc(rows*sizeof(int*));
for(i=0;i<rows;i++)
{
arr[i] = (int*)malloc(cols*sizeof(int));
}
// use new instead of malloc as using malloc leads to memory leaks
`enter code here
int **adj_list = new int*[rowsize];
for(int i = 0; i < rowsize; ++i)
{
adj_list[i] = new int[colsize];
}

How do I correctly set up, access, and free a multidimensional array in C?

I have seen dozens of questions about “what’s wrong with my code” regarding multidimensional arrays in C. For some reason people can’t seem to wrap their head around what is happening here, so I decided to answer this question as a reference to others:
How do I correctly set up, access, and free a multidimensional array in C?
If others have helpful advice, please feel free to post along!
In C since C99, even dynamic multidimensional arrays can be easily allocated in one go with malloc and freed with free:
double (*A)[n] = malloc(sizeof(double[n][n]));
for (size_t i = 0; i < n; ++i)
for (size_t j = 0; j < n; ++j)
A[i][j] = someinvolvedfunction(i, j);
free(A);
There are at least four different ways to create or simulate a multi-dimensional array in C89.
One is "allocate each row separately", described by Mike in his answer. It is not a multidimensional array, it merely imitates one (in particular it mimics the syntax for accessing an element). It can be useful in the case where each row has different size, so you aren't representing a matrix but rather something with a "ragged edge".
One is "allocate a multidimensional array". It looks likes this:
int (*rows)[NUM_ROWS][NUM_COLS] = malloc(sizeof *rows);
...
free(rows);
Then the syntax to access element [i,j] is (*rows)[i][j]. In C89, both NUM_COLS and NUM_ROWS must be known at compile-time. This is a true 2-D array, and rows is a pointer to it.
One is, "allocate an array of rows". It looks like this:
int (*rows)[NUM_COLS] = malloc(sizeof(*rows) * NUM_ROWS);
...
free(rows);
Then the syntax to access element [i,j] is rows[i][j]. In C89, NUM_COLS must be known at compile-time. This is a true 2-D array.
One is, "allocate a 1-d array and pretend". It looks like this:
int *matrix = malloc(sizeof(int) * NUM_COLS * NUM_ROWS);
...
free(matrix);
Then the syntax to access element [i,j] is matrix[NUM_COLS * i + j]. This (of course) is not a true 2-D array. In practice it has the same layout as one.
Statically speaking, this is easy to understand:
int mtx[3][2] = {{1, 2},
{2, 3},
{3, 4}};
Nothing complicated here. 3 rows, 2 columns; data in column one: 1, 2, 3; data in column two: 2, 3, 4.
We can access the elements via the same construct:
for(i = 0; i<3; i++){
for(j = 0; j<2; j++)
printf("%d ", mtx[i][j]);
printf("\n");
}
//output
//1 2
//2 3
//3 4
Now let’s look at this in terms of Pointers:
The brackets are a very nice construct to help simplify things, but it doesn’t help when we need to work in a dynamic environment, so we need to think of this in terms of pointers. If we want to store a “row” of integers, we need an array:
int row[2] = {1,2};
And you know what? We can access this just like a pointer.
printf("%d, %d\n",*row,*(row+1)); //prints 1, 2
printf("%d, %d\n",row[0],row[1]); //prints 1, 2
Now if we don’t know the number of values in a row we can make this array a dynamic length if we have a pointer to int, and we give it some memory:
int *row = malloc(X * sizeof(int)); //allow for X number of ints
*row = 1; //row[0] = 1
*(row+1) = 2; //row[1] = 2
…
*(row+(X-1)) = Y; // row[x-1] = Some value y
So now we have a dynamic 1 dimensional array; a single row. But we want lots of rows, not just one, and we don’t know how many. That means we need another dynamic 1 dimensional array, each element of that array will be a pointer which points to a row.
//we want enough memory to point to X number of rows
//each value stored there is a pointer to an integer
int ** matrix = malloc(X * sizeof(int *));
//conceptually:
(ptr to ptr to int) (pointer to int)
**matrix ------------> *row1 --------> [1][2]
*row2 --------> [2][3]
*row3 --------> [3][4]
Now all that’s left to do is to write the code which will perform these dynamic allocations:
int i, j, value = 0;
//allocate memory for the pointers to rows
int ** matrix = malloc(Rows * sizeof(int*));
//each row needs a dynamic number of elements
for(i=0; i<Rows; i++){
// so we need memory for the number of items in each row…
// we could call this number of columns as well
*(matrix + i) = malloc(X * sizeof(int));
//While we’re in here, if we have the items we can populate the matrix
for(j=0; j<X; j++)
*(*(matrix+i)+j) = value; // if you deference (matrix + i) you get the row
// if you add the column and deference again, you
// get the actual item to store (not a pointer!)
}
One of the most important things to do now is to make sure we free the memory when we’re done. Each level of malloc() should have the same number of free() calls, and the calls should be in a FILO order (reverse of the malloc calls):
for(i=0; i<Rows; i++)
free(*(matrix + i));
free(matrix);
//set to NULL to clean up, matrix points to allocated memory now so let’s not use it!
matrix = NULL;
If you want to use a typedef'd array, it is even simpler.
Say you have in your code typedef int LabeledAdjMatrix[SIZE][SIZE];
You can then use:
LabeledAdjMatrix *pMatrix = malloc(sizeof(LabeledAdjMatrix));
Then you can write:
for (i=0; i<SIZE; i++) {
for (j=0; j<SIZE; j++) (*parr)[i][j] = k++; /* or parr[0][i][j]... */
}
Because pArr is a pointer to you matrix and * has lower priority than [];
That is why a mode common idiom is to typedef the row:
typedef int LabeledAdjRow[SIZE];
Then you can write:
LabeledAdjRow *pMatrix = malloc(sizeof(LabeledAdjRow) * SIZE);
for (i=0; i<SIZE; i++) {
for (j=0; j<SIZE; j++) parr[i][j] = k++;
}
And you can memcpy all that directly:
LabeledAdjRow *pOther = malloc(sizeof(LabeledAdjRow) * SIZE);
memcpy(pOther, pMatrix, sizeof(LabeledAdjRow) * SIZE);
Going off of Jen's answer, if you want to allocate space for a 3D array, in the same manner, you can do
int(*A)[n][n] = malloc(sizeof(int[num_of_2D_arrays][n][n]));
for (size_t p = 0; p < num_of_2D_arrays; p++)
for (size_t i = 0; i < n; i++)
for (size_t j = 0; j < n; j++)
A[p][i][j] = p;
for (size_t p = 0; p < num_of_2D_arrays; p++)
printf("Outter set %lu\n", p);
for (size_t i = 0; i < n; i++)
for (size_t j = 0; j < n; j++)
printf(" %d", A[p][i][j]);
printf("\n");
free(A);

How do I work with dynamic multi-dimensional arrays in C?

Does someone know how I can use dynamically allocated multi-dimensional arrays using C? Is that possible?
Since C99, C has 2D arrays with dynamical bounds. If you want to avoid that such beast are allocated on the stack (which you should), you can allocate them easily in one go as the following
double (*A)[n] = malloc(sizeof(double[n][n]));
and that's it. You can then easily use it as you are used for 2D arrays with something like A[i][j]. And don't forget that one at the end
free(A);
Randy Meyers wrote series of articles explaining variable length arrays (VLAs).
With dynamic allocation, using malloc:
int** x;
x = malloc(dimension1_max * sizeof(*x));
for (int i = 0; i < dimension1_max; i++) {
x[i] = malloc(dimension2_max * sizeof(x[0]));
}
//Writing values
x[0..(dimension1_max-1)][0..(dimension2_max-1)] = Value;
[...]
for (int i = 0; i < dimension1_max; i++) {
free(x[i]);
}
free(x);
This allocates an 2D array of size dimension1_max * dimension2_max. So, for example, if you want a 640*480 array (f.e. pixels of an image), use dimension1_max = 640, dimension2_max = 480. You can then access the array using x[d1][d2] where d1 = 0..639, d2 = 0..479.
But a search on SO or Google also reveals other possibilities, for example in this SO question
Note that your array won't allocate a contiguous region of memory (640*480 bytes) in that case which could give problems with functions that assume this. So to get the array satisfy the condition, replace the malloc block above with this:
int** x;
int* temp;
x = malloc(dimension1_max * sizeof(*x));
temp = malloc(dimension1_max * dimension2_max * sizeof(x[0]));
for (int i = 0; i < dimension1_max; i++) {
x[i] = temp + (i * dimension2_max);
}
[...]
free(temp);
free(x);
Basics
Arrays in c are declared and accessed using the [] operator. So that
int ary1[5];
declares an array of 5 integers. Elements are numbered from zero so ary1[0] is the first element, and ary1[4] is the last element. Note1: There is no default initialization, so the memory occupied by the array may initially contain anything. Note2: ary1[5] accesses memory in an undefined state (which may not even be accessible to you), so don't do it!
Multi-dimensional arrays are implemented as an array of arrays (of arrays (of ... ) ). So
float ary2[3][5];
declares an array of 3 one-dimensional arrays of 5 floating point numbers each. Now ary2[0][0] is the first element of the first array, ary2[0][4] is the last element of the first array, and ary2[2][4] is the last element of the last array. The '89 standard requires this data to be contiguous (sec. A8.6.2 on page 216 of my K&R 2nd. ed.) but seems to be agnostic on padding.
Trying to go dynamic in more than one dimension
If you don't know the size of the array at compile time, you'll want to dynamically allocate the array. It is tempting to try
double *buf3;
buf3 = malloc(3*5*sizeof(double));
/* error checking goes here */
which should work if the compiler does not pad the allocation (stick extra space between the one-dimensional arrays). It might be safer to go with:
double *buf4;
buf4 = malloc(sizeof(double[3][5]));
/* error checking */
but either way the trick comes at dereferencing time. You can't write buf[i][j] because buf has the wrong type. Nor can you use
double **hdl4 = (double**)buf;
hdl4[2][3] = 0; /* Wrong! */
because the compiler expects hdl4 to be the address of an address of a double. Nor can you use double incomplete_ary4[][]; because this is an error;
So what can you do?
Do the row and column arithmetic yourself
Allocate and do the work in a function
Use an array of pointers (the mechanism qrdl is talking about)
Do the math yourself
Simply compute memory offset to each element like this:
for (i=0; i<3; ++i){
for(j=0; j<3; ++j){
buf3[i * 5 + j] = someValue(i,j); /* Don't need to worry about
padding in this case */
}
}
Allocate and do the work in a function
Define a function that takes the needed size as an argument and proceed as normal
void dary(int x, int y){
double ary4[x][y];
ary4[2][3] = 5;
}
Of course, in this case ary4 is a local variable and you can not return it: all the work with the array must be done in the function you call of in functions that it calls.
An array of pointers
Consider this:
double **hdl5 = malloc(3*sizeof(double*));
/* Error checking */
for (i=0; i<3; ++i){
hdl5[i] = malloc(5*sizeof(double))
/* Error checking */
}
Now hdl5 points to an array of pointers each of which points to an array of doubles. The cool bit is that you can use the two-dimensional array notation to access this structure---hdl5[0][2] gets the middle element of the first row---but this is none-the-less a different kind of object than a two-dimensional array declared by double ary[3][5];.
This structure is more flexible then a two dimensional array (because the rows need not be the same length), but accessing it will generally be slower and it requires more memory (you need a place to hold the intermediate pointers).
Note that since I haven't setup any guards you'll have to keep track of the size of all the arrays yourself.
Arithmetic
c provides no support for vector, matrix or tensor math, you'll have to implement it yourself, or bring in a library.
Multiplication by a scaler and addition and subtraction of arrays of the same rank are easy: just loop over the elements and perform the operation as you go. Inner products are similarly straight forward.
Outer products mean more loops.
If you know the number of columns at compile time, it's pretty simple:
#define COLS ...
...
size_t rows;
// get number of rows
T (*ap)[COLS] = malloc(sizeof *ap * rows); // ap is a *pointer to an array* of T
You can treat ap like any 2D array:
ap[i][j] = x;
When you're done you deallocate it as
free(ap);
If you don't know the number of columns at compile time, but you're working with a C99 compiler or a C2011 compiler that supports variable-length arrays, it's still pretty simple:
size_t rows;
size_t cols;
// get rows and cols
T (*ap)[cols] = malloc(sizeof *ap * rows);
...
ap[i][j] = x;
...
free(ap);
If you don't know the number of columns at compile time and you're working with a version of C that doesn't support variable-length arrays, then you'll need to do something different. If you need all of the elements to be allocated in a contiguous chunk (like a regular array), then you can allocate the memory as a 1D array, and compute a 1D offset:
size_t rows, cols;
// get rows and columns
T *ap = malloc(sizeof *ap * rows * cols);
...
ap[i * rows + j] = x;
...
free(ap);
If you don't need the memory to be contiguous, you can follow a two-step allocation method:
size_t rows, cols;
// get rows and cols
T **ap = malloc(sizeof *ap * rows);
if (ap)
{
size_t i = 0;
for (i = 0; i < cols; i++)
{
ap[i] = malloc(sizeof *ap[i] * cols);
}
}
ap[i][j] = x;
Since allocation was a two-step process, deallocation also needs to be a two-step process:
for (i = 0; i < cols; i++)
free(ap[i]);
free(ap);
malloc will do.
int rows = 20;
int cols = 20;
int *array;
array = malloc(rows * cols * sizeof(int));
Refer the below article for help:-
http://courses.cs.vt.edu/~cs2704/spring00/mcquain/Notes/4up/Managing2DArrays.pdf
Here is working code that defines a subroutine make_3d_array to allocate a multidimensional 3D array with N1, N2 and N3 elements in each dimension, and then populates it with random numbers. You can use the notation A[i][j][k] to access its elements.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
// Method to allocate a 2D array of floats
float*** make_3d_array(int nx, int ny, int nz) {
float*** arr;
int i,j;
arr = (float ***) malloc(nx*sizeof(float**));
for (i = 0; i < nx; i++) {
arr[i] = (float **) malloc(ny*sizeof(float*));
for(j = 0; j < ny; j++) {
arr[i][j] = (float *) malloc(nz * sizeof(float));
}
}
return arr;
}
int main(int argc, char *argv[])
{
int i, j, k;
size_t N1=10,N2=20,N3=5;
// allocates 3D array
float ***ran = make_3d_array(N1, N2, N3);
// initialize pseudo-random number generator
srand(time(NULL));
// populates the array with random numbers
for (i = 0; i < N1; i++){
for (j=0; j<N2; j++) {
for (k=0; k<N3; k++) {
ran[i][j][k] = ((float)rand()/(float)(RAND_MAX));
}
}
}
// prints values
for (i=0; i<N1; i++) {
for (j=0; j<N2; j++) {
for (k=0; k<N3; k++) {
printf("A[%d][%d][%d] = %f \n", i,j,k,ran[i][j][k]);
}
}
}
free(ran);
}
There's no way to allocate the whole thing in one go. Instead, create an array of pointers, then, for each pointer, create the memory for it. For example:
int** array;
array = (int**)malloc(sizeof(int*) * 50);
for(int i = 0; i < 50; i++)
array[i] = (int*)malloc(sizeof(int) * 50);
Of course, you can also declare the array as int* array[50] and skip the first malloc, but the second set is needed in order to dynamically allocate the required storage.
It is possible to hack a way to allocate it in a single step, but it would require a custom lookup function, but writing that in such a way that it will always work can be annoying. An example could be L(arr,x,y,max_x) arr[(y)*(max_x) + (x)], then malloc a block of 50*50 ints or whatever and access using that L macro, e.g.
#define L(arr,x,y,max_x) arr[(y)*(max_x) + (x)]
int dim_x = 50;
int dim_y = 50;
int* array = malloc(dim_x*dim_y*sizeof(int));
int foo = L(array, 4, 6, dim_x);
But that's much nastier unless you know the effects of what you're doing with the preprocessor macro.
int rows, columns;
/* initialize rows and columns to the desired value */
arr = (int**)malloc(rows*sizeof(int*));
for(i=0;i<rows;i++)
{
arr[i] = (int*)malloc(cols*sizeof(int));
}
// use new instead of malloc as using malloc leads to memory leaks
`enter code here
int **adj_list = new int*[rowsize];
for(int i = 0; i < rowsize; ++i)
{
adj_list[i] = new int[colsize];
}

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