I can't display values inside array

Question

I'm very new to using pointers and arrays, and I'm trying to write a program that, depending on your inputted choice, would execute a function that allows me to either define the array size and content, or display my array if there is one. But currently I can't seem to run my display function. When I try to execute case 2, it just immediately print out "invalid" and return to my menu. It doesn't seem like it's updating my array size and I don't know how to fix it.

#include <stdio.h>

#define MAX 100

void input(int arr[], int* size);
void show(int arr[], int size);
int i = 0

int main()
{
    int arr[MAX], size = 0, choice = 0;
    printf("Enter your choice: ");
    scanf("%d", &choice);
    switch(choice)
    {
        case 1:
            input(&arr, &size);
            break;
        case 2:
            if (size > 0)
            {
                show(arr, size);
            }
            else {
                printf("invalid");
            }
            break;
    }
}
void input(int arr[], int *size)
{
    int n;
    do {
        scanf("%d", &n);
        if (n < 0 || n > MAX)
        {
            printf("Invalid size, try again.\n");
            scanf("%d", &n);
        }
    }while (n < 0 || n > MAX);
    size = &n;
    for (i; i < n; i++)
    {
        printf("Input value for array %d: \n", i + 1);
        scanf("%d", &arr[i]);
    }
}

void show(int arr[], int size)
{
    for (i; i < size; i++)
    {
        printf("%d", arr[i]);
    }
}