Changing elements of an array by passing by reference to a function

Question

I am trying to pass an array by reference and change the values in the array in changeArray(). I am getting an error which states "Access violation writing location 0x00000001." I read Changing array inside function in C and I used Ryyker's answer to achieve my intended result (to have x[]={1,1,1,1,1]) but I get the aforementioned error. Here's my code:

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

int changeArray(int **a);

int main(void) {

    int *x[5] = { 1,5,4,3,1 };
    int *y[5] = { 1,5,4,3,1 };

    changeArray(&x);
    for (int z = 0; z <= 4; ++z) {
        printf_s("%s", x[z]);
    }
    free(x);
}

int changeArray(int **a) {

    for (int z = 0; z < 5; ++z) {
        (*a)[z] = 1;
    }
}

I know there are similar posts, but all the ones I have seen don't seem to solve my problem, any help is appreciated!

`int *x[5]` should be `int x[5]` (the same for `y`) and `int changeArray(int **a)` should be `int changeArray(int (*a)[5])`. — mch, Oct 25 '18 at 06:41
Remove the * before x and y. Also, you shouldn't call `free(x)`, as the array is not heap allocated — Petter Hesselberg, Oct 25 '18 at 06:42
With this statement you are filling address 1,5,4,3,1 in pointer x[0], x[1] ... int *x[5] = { 1,5,4,3,1 }; It is similar like int *x = 5; which is not valid better take integer type array it does not mandatory variable to be of pointer type for call by reference. — anshkun, Oct 25 '18 at 07:01
How did you **not** get any compiler diagnostics? There are *none* mentioned in your post. Please find out where the compiler outputs the diagnostics messages! — Antti Haapala -- Слава Україні, Oct 25 '18 at 07:02
`"Access violation writing location 0x00000001."` **Ouch!**. You are trying to access memory at Address `1` which is at the very bottom of the *System Reserved* memory pool. Not good. — David C. Rankin, Oct 25 '18 at 07:11

score 3 · Answer 1 · answered Oct 25 '18 at 06:46

3

Your program is doing something completely unintended.

int *x[5] = { 1,5,4,3,1 };
int *y[5]= { 1,5,4,3,1 };

Here you're initializing a bunch of int pointers with values from 1 to 5. So they point to invalid memory.

Then later, here:

for (int z = 0; z <= 4; ++z) {
    printf_s("%s", x[z]);
}

You are telling it to print the string at that invalid memory.

The 0x00000001 in Access violation writing location 0x00000001 is actually the hex representation of the first 1 in int *x[5] = { 1,5,4,3,1 };.

What you probably want is this:

int changeArray(int *a) {

    for (int z = 0; z < 5; ++z) {
        a[z] = 1;
    }
}

And this:

int main(void) {

    int x[5] = { 1,5,4,3,1 };

    changeArray(x);
    for (int z = 0; z <= 4; ++z) {
        printf("%d", x[z]); // also consider adding space, such as "%d "
    }
}

answered Oct 25 '18 at 06:46

Blaze

16,736
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1

Well done, there was no need to pass a pointer to array to begin with since the array had automatic storage type with its scope in `main()` to begin with... – David C. Rankin Oct 25 '18 at 07:13
Yeah, I think OP misundestood how pointers and arrays work in C. I can relate, it took me a good while to understand those properly. – Blaze Oct 25 '18 at 07:21
1

We all have that same bump on our foreheads from beating our head into that same wall when we were learning `:)` – David C. Rankin Oct 25 '18 at 07:21

mch · Answer 2 · 2018-10-25T06:58:11.830

1

int *x[5] should be int x[5].
You do not need int y[5] at all.
int changeArray(int **a) should be void changeArray(int (*a)[5]).
You pass a pointer to an array and return nothing. free(x); is wrong, x is on the stack and must not be freed.
printf_s("%s", x[z]); should be printf("%d ", x[z]);, x[z] is an int and so it needs %d as format specifier. Also note the space after it to see the different numbers and not only a large number.

Here is your corrected code https://ideone.com/kwXrFg

edited Oct 25 '18 at 06:58

answered Oct 25 '18 at 06:46

mch

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Nice answer mch, but it's not like `x` doesn't need to be freed, it must not be freed, as I mention in my answer. – gsamaras Oct 25 '18 at 06:54

gsamaras · Answer 3 · 2018-10-25T06:53:23.507

The code should be like this:

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

void changeArray(int (*a)[5]);

int main(void) {
    int x[5] = { 1,5,4,3,1 };
    changeArray(&x);
    for (int z = 0; z <= 4; ++z) {
        printf_s("%d", x[z]);
    }
    return 0;
}

void changeArray(int (*a)[5]) {
    for (int z = 0; z < 5; ++z) {
        (*a)[z] = 1;
    }
}

and give output:

11111

as you can see in the Live Demo.

Here are the changes I made:

Change int *x[5] = { 1,5,4,3,1 }; to int x[5] = { 1,5,4,3,1 };.
Remove y, since you do not use it.
Change the prototype of your function to: void changeArray(int (*a)[5]);, since you don't return anything, and the parameter is also changed, to pass array x as is with the changes now.
Use %d to print integers, not %s.
Remove free(x), since you don't dynamic allocate memory for array x, thus you must not de-allocate it manually.

Thanks gsamaras this is interesting :) – anshkun Oct 25 '18 at 06:58 — anshkun, Oct 25 '18 at 06:58

score 0 · Answer 4 · edited Oct 25 '18 at 07:04

0

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

int changeArray(int *a);

int main(void) {

    int x[5] = { 1,5,4,3,1 };
    int y[5]= { 1,5,4,3,1 };

    changeArray(x);
    for (int z = 0; z <= 4; ++z) {
        printf_s("%s", x[z]);
    }
}

int changeArray(int *a) {

    for (int z = 0; z < 5; ++z) {
        a[z] = 1;
    }
}

edited Oct 25 '18 at 07:04

mch

9,424
2
28
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answered Oct 25 '18 at 07:03

user1888149

143
1
4

Changing elements of an array by passing by reference to a function

4 Answers4