How to Pass Array to function in C

C code to pass array to a function
#include<stdio.h>
#include<stdlib.h>
void func(int *arr, int n);
int main()
{
int i;
int arr[]={1,2,3,4,5};

    printf("\nArray element after function : \n");
func(arr,5);//its a pass by reference only as name of the arry is the base address
    for(i=0;i<5;i++)
    {
        printf("%d ",arr[i]);//11 12 13 14 15
    }
    printf("\n");
    return 0;
}
void func(int *p, int n)
{
int *arr=p;
    int i;
    for(i=0;i<n;i++)
    {
        arr[i]=p[i]+10;
    }

}//end of main

Example 2:
/*
   c code to pass array to a function
 */
#include<stdio.h>
#include<stdlib.h>
int* func(int *arr, int n);
int main()
{
int i;
int arr[]={1,2,3,4,5};
int *p;
p=func(arr,5);
    printf("\nArray element after function : \n");
    for(i=0;i<5;i++)
    {
        printf("%d ",p[i]);//11,12,13,14,15
    }
    printf("\n");
    return 0;
}
int* func(int *arr, int n)
{
    int i;
    int *p=arr;
    p=(int *)malloc(n*sizeof(int));
    for(i=0;i<n;i++)
    {
        p[i]=arr[i]+10;//arr[i] is must not p[i]
    }
    return p;
}

Example 3: not a good program
/*
   c code to pass array to a function
 */
#include<stdio.h>
#include<stdlib.h>
int* func(int *arr, int n);
int main()
{
    int i;
    int arr[]={1,2,3,4,5};

    printf("\nArray element after function : \n");
    func(arr,5);
    for(i=0;i<5;i++)
    {
        printf("%d ",arr[i]);//1 2 3 4 5
    }
    printf("\n");
    return 0;
}
int*  func(int *p, int n)
{
    int *arr=p;
    arr=(int *)malloc(n*sizeof(int));
    int i;
    for(i=0;i<n;i++)
    {
        arr[i]=arr[i]+10;//arr[i] is must not p[i]
    }
    return arr;/*if returning a pointer then function should have been called from a pointer */

}

Bugs in pointer commonly done in C

/* code for showing: most common error in pointer is assigning a value()not address to pointer without mallocing it. Before this note the declaration like int *ptr; tells the compiler that it will store address of an integer and not a value directly*/
#include<stdio.h>
#include<stdlib.h>
main()
{
    /*case when pointer need to malloced:when ptr is directly assigned with VALUE and not address*/
    int *p;//p is pointer so should store address and not value.
    //  *p=10;///Ptr Error: assigning value(not address) to pointer mallocing
    //printf("%d\n",*p);//as a result segfault


    /*so allocate memory for ptr then assign valut to pointer*/
    p=malloc(1*sizeof(int));
    *p=10;
    printf("%d\n",*p);

    /*Error not freeing memory allocated dynamically cause mem leak*/
    free(p);

    /*case when pointer need not to malloced: when ptr is assigned with ADDRESS and not vakue*/
    int *q;//this statement tells the compiler that q will store address of an integer.
    int a=10;
    q=&a;//no allocation required for q when address is assigned to pointer
    printf("a=%d\n\n",*q);//some times work but not always without malloc
}

Also understand well declaration of variables and pointers very well to avoid errors/bug:
int i=20;//This declaration tells three things to compiler. reserve memory for a int size, name it i and store 20 there.
int *p;//tells the compiler that p will be used to store base address of an integer.

char *p;//tells the compiler that p will be used to store base address of a string.
char *ptr="Ram"; is valid as ptr is assigned with the base address of string ie char array since name of an array gives the base address and not value.
int *p=10;//will be an array as here pointer is assigned a value and not address like above statement.
Solution in this case will be to first allocate memory then assign value.
*so allocate memory for ptr then assign valut to pointer*/
    p=malloc(1*sizeof(int));
    *p=10;
    printf("%d\n",*p);