How to use pointer type in C?

Question

I am recently studying pointers and arrays. Suppose I initialize an array like

 int a[4]={6,2,3,4};

Now after reading a lot ,I understand that

1) a and &a will point to same location.But they aren't pointing to the same type.

2) a points to the first element of the array which is an integer value so the type of a is int*.

3) &a points to an entire array(i.e an integer array of size 4) therefore the type is int (*)[].

Now what I actually don't get is how to use these type??

For Example: CODE 1:

#include<stdio.h>
void foo(int (*arr)[4],int n)
{
    int i;
    for(i=0;i<n;i++)
        printf("%d ",*(*arr+i));
    printf("\n");
}
int main()
{
    int arr[4]={6,2,3,4};
    foo(&arr,4);
    return 0;
}

CODE 2:

#include<stdio.h>
void foo(int *arr,int n)
{
    int i;
    for(i=0;i<n;i++)
        printf("%d ",*(arr+i));
    printf("\n");
}
int main()
{
    int arr[4]={6,2,3,4};
    foo(&arr,4);
    return 0;
}

In code 2 we are passing &arr so its type should be of int (*)[],then how come we are getting the correct output even though we are having a type int *.

I really don't understand what is the meaning of type and how to use it?

Kindly explain with some examples.

Answer 1

The best way to pass arrays in to functions is a pointer and the number of elements.

Note you can pass arr in to the function as arr or &arr[0]; both resolve to the same address.

#include<stdio.h>
void foo(int *arr,int n)
{
    int i;
    for(i=0;i<n;i++)
        printf("%d ",arr[i]);
    printf("\n");
}
int main()
{
    int arr[4]={6,2,3,4};
    foo(arr,4);
    return 0;
}

Also, it's better to index the array, rather than addition and dereference. e.g. *(arr+i) becomes arr[i].

Answer 2

Code 2 is not valid C, because a conversion between an array pointer and a pointer to int is not a valid pointer conversion - they are not compatible types. If it works, it is because of some non-standard extension of your compiler, or possibly you just "got lucky".

Please note that the best way to pass an array to a function in modern C is this:

void foo(int n, int arr[n])
{
    for(int i=0;i<n;i++)
        printf("%d ",arr[i]);
    printf("\n");
}

As part of a function parameter list, arr will get adjusted to a pointer to the first element of the array, type int*.

The pedantically correct version would also replace int n and int i with size_t, which is the most proper type to use when describing the size of an object.