Free first element of array

Question

When I allocate an array using malloc, is there a way to only free the first element(s) of the array?

A small example:

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

int main() {
    char * a = malloc(sizeof(char) * 8);
    strcpy(a, "foo bar");

    // How I would have to do this.
    char * b = malloc(sizeof(char) * 7);
    strcpy(b, a+1);


    free(a);
    free(b);
}

Is there a way to free just the first char of a, so that I can use the rest of the string using a+1?

No you cant free only one byte of array. You can free only pointer returned from `malloc`, which points on the first byte of allocated array, and as well, it will free all the allocated array of bytes. Anyway you can realloc it to size you want, which can be more or less bytes then actuall size. — kocica, Aug 18 '17 at 13:52
Take a look at [this](https://stackoverflow.com/a/27876929/2173917) Dupe??? — Sourav Ghosh, Aug 18 '17 at 13:54
You could work with another pointer, with `char *b = a + 1;` and later on, you just `free(a);` — Weather Vane, Aug 18 '17 at 14:05
@WeatherVane I know. I was just wondering if that would be possible. — Ward Segers, Aug 18 '17 at 14:08

ad absurdum · Accepted Answer · 2017-08-18T14:57:40.500

If you want to remove the first character of a, you could use memmove() to move the remainder of the characters in the string to the left by 1, and you could use realloc() to shrink the allocation if desired:

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

int main(void)
{
    char * a = malloc(sizeof(char) * 8);
    strcpy(a, "foo bar");

    puts(a);

    size_t rest = strlen(a);

    memmove(a, a+1, rest);

    /* If you must reallocate */
    char *temp = realloc(a, rest);
    if (temp == NULL) {
        perror("Unable to reallocate");
        exit(EXIT_FAILURE);
    }
    a = temp;

    puts(a);

    free(a);

    return 0;
}

Update

@chux has made a couple of good points in the comments.

First, instead of exiting on a failure in realloc(), it may be better to simply continue without reassigning temp to a; after all, a does point to the expected string anyway, the allocated memory would just be a little larger than necessary.

Second, if the input string is empty, then rest will be 0. This leads to problems with realloc(a, rest). One solution would be to check for rest == 0 before modifying the string pointed to by a.

Here is a slightly more general version of the above code that incorporates these suggestions:

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

int main(void)
{
    char *s = "foo bar";
    char *a = malloc(sizeof *a * (strlen(s) + 1));
    strcpy(a, s);

    puts(a);

    size_t rest = strlen(a);

    /* Don't do anything if a is an empty string */
    if (rest) {
        memmove(a, a+1, rest);

        /* If you must reallocate */
        char *temp = realloc(a, rest);
        if (temp) {
            a = temp;
        }
    }

    puts(a);

    free(a);

    return 0;
}

Note: This is a case where code could use `if (temp) { a = temp; }` for on a rare failure of shrinking an allocation, simply enough to not update `a`. — chux - Reinstate Monica, Aug 18 '17 at 14:30
Note: Additional considerations needed when `rest == 0`. OP's intent is unclear in this case. — chux - Reinstate Monica, Aug 18 '17 at 14:32
@chux-- those are both good points; I have added an update to my answer to incorporate your suggestions. — ad absurdum, Aug 18 '17 at 14:58
A downvote with comment may be helpful; a downvote without comment (on a post that is three months old, no less) is just mysterious. — ad absurdum, Nov 28 '17 at 18:13

haccks · Answer 2 · 2017-08-18T15:20:04.920

0

Is there a way to free just the first char of a

No. You can not free the first character of a because it is of char type. Only pointer returned by malloc family function can be freed. You can do this though.

char * a = malloc(sizeof(char) * 8);
strcpy(a, "foo bar");
char * b = malloc(strlen(a));
strcpy(b, a+1);
free(a);

edited Aug 18 '17 at 15:20

answered Aug 18 '17 at 14:56

haccks

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Free first element of array

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