C array manipulation in function

Question

In this code i want to user to input numbers in the function then print the array in the main method, but when i try to print in the main method it gives me random numbers.

for eg: 1 2 3 gives 1 3 1895586112

void arrayInput(int *arr) {
    int x;
    int y;
    int z;
    printf("enter 3 numbers =");
    scanf("%d %d %d", &x,&y,&z);
    arr[0]=x;
    arr[1]=y;
    arr[2]=z;
}

int main(int argc, char **argv){
    int *arr[3];
    arrayInput(&arr);
    int i;
    for(i=0; i<3; i++)  
    {  
        printf("%d ", arr[i]);  
    } 
}

I want to change the array values without changing the method or param type

Just pay attention to your compiler. The code you have posted isn't valid C and the compiler told you as much. If you are a beginner and using gcc then compile with something like -Wall -Wextra -Werror. — Lundin
– Lundin, Commented Apr 7, 2021 at 7:30

Jabberwocky · Accepted Answer · 2021-04-07 06:58:22Z

1

int *arr[3] is an array of pointers to int. But you want an array of int.

Simply change:

int *arr[3];
arrayInput(&arr);

to

int arr[3];
arrayInput(arr);

This should be covered in your C text book.

Bonus:

You can remplace this convoluted code:

void arrayInput(int *arr) {
    int x;
    int y;
    int z;
    printf("enter 3 numbers =");
    scanf("%d %d %d", &x,&y,&z);
    arr[0]=x;
    arr[1]=y;
    arr[2]=z;
}

with this:

void arrayInput(int *arr) {
    printf("enter 3 numbers =");
    scanf("%d %d %d", &arr[0],&arr[1], &arr[2]); 
}

answered Apr 7, 2021 at 6:58

Jabberwocky

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3 Comments

David C. Rankin Over a year ago

Better int arrayInput(int *arr) { ... return scanf("%d %d %d", &arr[0],&arr[1], &arr[2]) == 3; }

Jabberwocky Over a year ago

@0___________ scanf isn't fit for user input anyway, but using it without error checks for toy programs is OK IMO.

danny19 Over a year ago

Thank you, this seems to work perfectly. Also the bonus was helpful too :)

quantummind · Accepted Answer · 2021-04-07 08:10:22Z

1

There are multiple issues with your code. First you have declared int *arr[3]; which would mean pointer arrays. For your task a simple array is needed. The array variable contains the address of the first element, so you don't need &. A simple corrected code is as follows:

#include <stdio.h>

void arrayInput(int arr[]) {
    int x;
    int y;
    int z;
    printf("enter 3 numbers =");
    scanf("%d %d %d", &x,&y,&z);
    arr[0]=x;
    arr[1]=y;
    arr[2]=z;
}

int main(int argc, char **argv){
    int arr[3];
    arrayInput(arr);
    int i;
    for(i=0; i<3; i++)  
    {  
        printf("%d ", arr[i]);  
    } 
}

edited Apr 7, 2021 at 8:10

answered Apr 7, 2021 at 6:59

quantummind

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3 Comments

David C. Rankin Over a year ago

"which would mean pointer to arrays" NO it means an array of 3 pointers to int. A pointer to array of int[3] would be int (*arr)[3]; Much much different. An array of pointers is just that, an array where the elements are pointers. A pointer to array is a single-pointer to an array type. (but I knew what you meant and your code correction is correct, but the language may leave the OP with the wrong impression of what a pointer to array is)

quantummind Over a year ago

Yes, you are right, I had a 'to' as mistake.

David C. Rankin Over a year ago

Much better. Not like I've never had a stray "to" or other misnomer in my answers :)

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