CODEWITHSUNDEEP
arrayscfunctionmethods
danny19
danny19

Reputation: 13

C array manipulation in function

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

Upvotes: 1

Views: 279

Answers (2)

quantummind
quantummind

Reputation: 2136

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]);  
    } 
}

Upvotes: 1

Jabberwocky
Jabberwocky

Reputation: 50774

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]); 
}

Upvotes: 1

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