Passing dynamically allocated integer arrays in C
I read an example on passing multidimensional arrays in C "on this site.
This is a great example using char arrays and I learned a lot from it. I would like to do the same by creating a function to handle a dynamically allocated one dimensional integer array and then create another function to handle a multidimensional integer array. I know how to do this as the return value of a function. But in this application, I need to do it in the argument list of the function.
As in the example above, I would like to pass a pointer to an integer array to the function, as well as the number of "num" elements (or "row" and "col" for a 2D array function, etc.). I got a reworked version of another example here, but I can't seem to get this to work, try it as best I can (lines of code that are new or changed from this example are marked). Does anyone know how to solve this?
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ELEMENTS 5
void make(char **array, int **arrayInt, int *array_size) {
int i;
char *t = "Hello, World!";
int s = 10; // new
array = malloc(ELEMENTS * sizeof(char *));
*arrayInt = malloc(ELEMENTS * sizeof(int *)); // new
for (i = 0; i < ELEMENTS; ++i) {
array[i] = malloc(strlen(t) + 1 * sizeof(char));
array[i] = StrDup(t);
arrayInt[i] = malloc( sizeof(int)); // new
*arrayInt[i] = i * s; // new
}
}
int main(int argc, char **argv) {
char **array;
int *arrayInt1D; // new
int size;
int i;
make(array, &arrayInt1D, &size); // mod
for (i = 0; i < size; ++i) {
printf("%s and %d\n", array[i], arrayInt1D[i]); // mod
}
return 0;
}
There are quite a few problems with this code. Take a look at the following:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ELEMENTS 5
/*
* A string is an array of characters, say char c[]. Since we will be creating
* an array of those, that becomes char *(c[]). And since we want to store the
* memory we allocate somewhere, we must be given a pointer. Hence char
* **(c[]).
*
* An int doesn't require a complete array, just int i. An array of those is
* int i[]. A pointer to those is then int *(i[]).
*/
void
make(char **(chars[]), int *(ints[]), size_t len)
{
static char hw[] = "Hello, World!";
size_t i = 0;
/*
* Allocate the memory required to store the addresses of len char arrays.
* And allocate the memory required to store len ints.
*/
*chars = malloc(len * sizeof(char *));
*ints = malloc(len * sizeof(int));
/* Fill each element in the array... */
for (i = 0; i < ELEMENTS; i++) {
/* ... with a *new copy* of "Hello world". strdup calls malloc under
* the hood! */
(*chars)[i] = strdup(hw);
/* ...with a multiple of 10. */
(*ints)[i] = i * 10;
}
}
int
main(void)
{
/* A string c is a character array, hence char c[] or equivalently char *c.
* We want an array of those, hence char **c. */
char **chars = NULL;
/* An array of ints. */
int *ints = NULL;
size_t i = 0;
/* Pass *the addresses* of the chars and ints arrays, so that they can be
* initialized. */
make(&chars, &ints, ELEMENTS);
for (i = 0; i < ELEMENTS; ++i) {
printf("%s and %d\n", chars[i], ints[i]);
/* Don't forget to free the memory allocated by strdup. */
free(chars[i]);
}
/* Free the arrays themselves. */
free(ints);
free(chars);
return EXIT_SUCCESS;
}
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The line size is missing here:
arrayInt[i] = malloc( sizeof(int)); // new
should look something like this:
arrayInt[i] = malloc( row_len * sizeof(int)); // new
where before you used the length of the given string, since the size of the string ( strlen(t)+1
must also be in parentheses, although the effect is the same as sizeof(char)
being 1)
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