通过传递给函数来更改数组的元素

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!

Answer 1

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 "
    }
}

Answer 2

• 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

Answer 3

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.

Answer 4

#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;
    }
}