Creating and looping a two-dimensional array from dynamic values in C - c

Question

I am trying to create a two dimensional array based on two pre-calculated values (here len1 and len2 are const, but they come from another function). When I run this I get segmentation fault. I am very new to C, this is my first task. Can not figure it out from Guides nor SO, anybody around to help me out?
I suppose the dynamic creation of the two dimensional arrays is wrong. But can't find a good example that would work ..
int main() {
int y, x;
int my_val = 10; // dynamnic value calculated by another func
int len1 = 3; // dynamnic value calculated by another func
int len2 = 3; // dynamnic value calculated by another func
int cols = len1 + 1;
int rows = len2 + 1;
int **twodarr = (int **)malloc(rows * cols * sizeof(int));
for (x = 1; x < cols; x++) {
for (y = 1; y < rows; y++) {
twodarr[y][x] = my_val;
}
}
return 0;
}

Answer 1

Arrays have never been first class elements in C, and multi-dimensional ones have even more poor support. Originally, only C used constant sized arrays, because pointer arithmetics was enough for dynamic 1D arrays, and pointers were an essential element of the language.
C99 introduced the concept of Variable Length Arrays which are what #Lundin's answer uses. Unfortunately, C11 defined them as an optional feature, and Microsoft choosed not to support them for compatibility with C++.
If you use a Microsoft compiler or want compability with environments that do not support the optional VLA feature, you will have to use the old linear idiom: you only use 1D arrays and use compound indices computation: the index of element (i, j) is j + i * cols where cols is the size of the second dimension.
Your code could become:
...
int *twodarr = malloc(rows * cols * sizeof(int)); // BEWARE: actualy 1D array!
for (x = 1; x < cols; x++) {
for (y = 1; y < rows; y++) {
twodarr[x + y*cols] = my_val;
}
}
...

Answer 2

You have to allocate each rows
// allocation of cols
int **twodarr = (int **)malloc(cols * sizeof(int*));// note it is sizeof(int*)
// allocation each rows (in each cols)
for (x = 0; x < cols; x++) {
twodarr[x] = (int *)malloc(rows * sizeof(int));
}

Answer 3

The problem is that int **twodarr cannot be used for 2D arrays, it has no relation what-so-ever to them. You need to swap it for a pointer to a 2D array. Or more conveniently, a pointer to a 1D array - a pointer to a row in this case, assuming [rows][cols].
Also, arrays in C start at index 0.
Code with bug fixes & a simple print example:
#include <stdio.h>
#include <stdlib.h>
int main() {
int my_val = 10; // dynamnic value calculated by another func
int len1 = 3; // dynamnic value calculated by another func
int len2 = 3; // dynamnic value calculated by another func
int rows = len2 + 1;
int cols = len1 + 1;
int (*twodarr)[cols] = malloc( sizeof(int[rows][cols]) );
for (int r = 0; r < rows; r++) {
for (int c = 0; c < cols; c++) {
twodarr[r][c] = my_val;
printf("%d ", twodarr[r][c]);
}
puts("");
}
free(twodarr);
return 0;
}

Answer 4

#kcabus had it right...and admittedly the much more readable way for sanity sake.
The other way to go about it would be to declare it as a memory block, but its much more confusing.
such as
int *twodarr = (int*)calloc((rows *
cols), sizeof(int));
// accessed as follows
*(twodarr + rows*r + c) = value;
// rows * position + position 2
// much more confusing.
A third alternative would be to create a struct like POINT (or just use point) and use two values by just creating an array of POINT just as an example. But I assume you don't want to deal with that in a loop...and I don't blame you heh.