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This article will demonstrate various ways to dynamically allocate arrays in C.

mallocDynamically allocate arrays in C using function

mallocThe allocate_by function is the core function for allocating dynamic memory on the heap. It allocates a given number of bytes and returns a pointer to the memory area. Therefore, if you want to dynamically allocate an array of some object type, you should first declare a pointer of that type. Next, you should call it by passing the number of elements multiplied by the size of a single object as a parameter malloc.

In the following example, we allocate memory to store a character string. As required by secure coding standards, errnoset to 0 and check mallocthe pointer returned by the call to verify successful execution of the function. Finally, memmovethe function is used to copy the string to the allocated memory location.

#include <stdlib.h>
#include <stdio.h>
#include <errno.h>
#include <string.h>

#define SIZE 100

const char *str = "random string to be moved";

int main() {
    char *arr = NULL;

    errno = 0;
    arr = malloc(SIZE * sizeof(char));
    if (!arr) {
        perror("malloc");
        exit(EXIT_FAILURE);
    }

    memmove(arr, str, strlen(str));
    printf("arr: %s\n", arr);

    free(arr);
    exit(EXIT_SUCCESS);
}

Output:

arr: random string to be moved

Use reallocthe function to modify the allocated memory area in C language

reallocThe getsize() function is used to modify the size of a memory area previously mallocallocated by realloca call to getsize(). It takes the original memory address and the new size as the second argument. Note that it may return the same pointer as passed, or it may return a different pointer depending on the requested size and the available memory after the given address. At the same time, the contents of the previous array will remain unchanged up to the newly specified size.

#include <stdlib.h>
#include <stdio.h>
#include <errno.h>
#include <string.h>

#define SIZE 100

const char *str = "random string to be moved";

int main() {
    char *arr = NULL;

    errno = 0;
    arr = malloc(SIZE);
    if (!arr) {
        perror("malloc");
        exit(EXIT_FAILURE);
    }

    int num = 102;  // User Provided Value
    for (int i = 0; i < num; ++i) {
        if (i > SIZE) {
            arr = realloc(arr, 2 * SIZE);

            if (!arr) {
                perror("realloc");
                exit(EXIT_FAILURE);
            }
        }

        arr[i] = 'a';
    }

    free(arr);
    exit(EXIT_SUCCESS);
}

Using macros to implement array allocation of given objects in C

Typically, mallocit is used to allocate arrays of some user-defined structures. Since mallocthe returned pointer is voida pointer and can be implicitly converted to any other type, it is better to explicitly convert the returned pointer to the corresponding type. Because it is easy to miss something and not include the correct markup, we implemented a macro expression that takes the number of elements in the array and the object type to automatically construct the correct mallocstatement, including the correct cast.

#include <stdlib.h>
#include <stdio.h>
#include <errno.h>
#include <string.h>

#define SIZE 100

typedef enum { Jan, Feb, MAR, APR, MAY, JUN, JUL, AUG, SEP, OCT, NOV, DEC} month;

typedef struct {
    unsigned char dd;
    month mm;
    unsigned yy;
} date;

#define MALLOC_ARRAY(number, type) \
    ((type *)malloc((number) * sizeof(type)))

int main() {
    date *d = NULL;

    errno = 0;
    d = MALLOC_ARRAY(SIZE, date);
    if (!d) {
        perror("malloc");
        exit(EXIT_FAILURE);
    }

    free(d);
    exit(EXIT_SUCCESS);
}