How to define matrix dynamically in C

Question

I created a program that prints 2 dimension matrix from external txt file. My program works if I define static matrix such as A[10][10]. But I want to allocate memory dynamically.

When I try this code:

int **A = (int **)malloc(N * N * sizeof(int));

There is error as following:

Unhandled exception at 0x00AC159B in dataStructures4.exe: 0xC0000005: Access violation writing location 0xCDCDCDCD.

On this loop:

for(i=0; i<N; i++){
    for(j=0; j<N; j++){
        A[i][j] = -1;
    }
}

I think I can't generate dynamic matrix correctly. How should I modify my code?

Answer 1

You dynamically create 2d arrays as a pointer to chunk of int* pointers which point to a chunk of ints.

So there are two steps:

1) A to point to the first of a chunk of int*

int **A = (int **)malloc(N * sizeof(int*));

2) each of them to point to the first of a chunk of ints

for(i=0; i<N; i++){
    A[i] = (int *)malloc(N * sizeof(int));
}

Answer 2

Use a pointer to an array of specific size:

You can use #define to define N, or use a variable:

int n = 10 ;
int (*A)[n] = malloc(n * n * sizeof(int));

This way you get a contiguous block of memory.

Answer 3

Remember that A[i][j] is the same as *(*(A+i)+j). Your code is broken because *(A+i) dereferences an invalid, uninitialized pointer. This happens on this line:

A[i][j] = -1;

The alternative is to use a dynamically allocated array of pointers, where each element points to another dynamically allocated array. You can do it like this:

int **array = malloc(N*sizeof(*array));
for (int i = 0; i < N; i++)
    array[i] = malloc(N*sizeof(*array[i]));

/* Use array[i][j]... */

Note that the memory layout is very different from that of a 2D array, but the expression array[i][j] will do what you want - it will "look" like you have a 2D array.

Answer 4

Allocate each row separately, like here:

int **A = (int **)malloc(N * sizeof(int*));

for(i=0; i<N; i++){
    A[i] = (int *)malloc(N * sizeof(int));
}

Answer 5

Same as an array, a 2 dimension matrix holds a contiguous address space. So theoretically there is a way to achieve this.

int m = 2;
int n = 4;
int *a = malloc(m*n*sizeof(int));
int **b = malloc(m*sizeof(int*));
for(int i=0, i<m, i++)
{
    b[i] = a+i*n;
}

In this manner, b can be used as a matrix as you demand. However, it will waste malloc(m*sizeof(int*)) to hold b.

or you can:

int m = 2;
int n = 4;
int *a = malloc(m*n*sizeof(int));

use a[m*i+j] as a[i][j]

Answer 6

Hope the below stuff is good. I haven't tested it, just providing you with syntax.

int** abc; // Your 2D array pointer abc[N][M]

abc = (int**)malloc(dim1_max * sizeof(int*)); // dim1_max is N
for (int i = 0; i < dim1_max; i++) {
  abc[i] = (int*)malloc(dim2_max * sizeof(int)); // dim2_max is M
}

[....your array used....]
//Once you allocate dynamically you need to free the memory. IMP point.
for (int i = 0; i < dim1_max; i++) {
  free(abc[i]);
}
free(abc);