Pointers in C while using array and swap function - c

Question

I had a doubt in using pointers
For example I made this swapping program and to swap so i have to use pointers in order to maintain same address where swapping is happening
#include<stdio.h>
int swap(int *x,int *y); //swapping function
int main()
{
int a,b;
scanf("%d %d",&a,&b);
printf("A->%d,B->%d\n",a,b); //print original array
swap(&a,&b); //pass by reference
printf("A->%d,B->%d\n",a,b); //print sorted array
}
int swap(int *x,int *y)
{
int temp; // Swapping
temp =*x;
*x=*y;
*y=temp;
}
Then I wrote a program for merge sort. Here I just passed the value (pass by value) in the funtions without using pointers but when I print the sorted array in the main function it is sorted.
#include<stdio.h>
void merge(int a[],int start,int mid,int end); //Sorting and merging lists
void mergesort(int a[],int start,int end); //Dividing the list
int main()
{
int size,i;
scanf("%d",&size);
int arr[size];
for(i=0;i<size;i++) //Input array
{
scanf("%d",&arr[i]);
}
printf("Array is: ");
for(i=0;i<size;i++)
{
printf("%d ",arr[i]); //print original array
}
printf("\n");
mergesort(arr,0,size-1); //Mergesort call-pass by value
printf("Sorted Array is: ");
for(i=0;i<size;i++)
{
printf("%d ",arr[i]); //print sorted array
}
printf("\n");
}
void mergesort(int a[],int start,int end)
{
int mid;
if(start<end)
{
mid=(start+end)/2; //calculating mid and dividing array recursively
mergesort(a,start,mid);
mergesort(a,mid+1,end);
merge(a,start,mid+1,end);
}
}
void merge(int a[],int start,int mid,int end)
{
int b[end]; //auxillary array
int i;
int p=start,q=mid,k=start;
while((p<mid) && (q<=end)) //comparing both list
{
if(a[p]<=a[q])
{
b[k++]=a[p++];
}
else
{
b[k++]=a[q++];
}
}
while(p<mid) //adding remaing items to auxillary array
{
b[k++]=a[p++];
}
while(q<=end) //adding remaing items to auxillary array
{
b[k++]=a[q++];
}
for(i=start;i<k;i++) //copying items from auxillary array to original array
{
a[i]=b[i];
}
}
Why is this happening ?

Answer 1

You can't directly pass arrays by value in C (best you can do is wrap a fixed-size array in a struct). In a parameter list int a[] is exactly equivalent to int *a.
So you are using pointers and that's why the array is sorted.

Answer 2

In C and similar languages the first array dimension of a function parameter is rewriten as a pointer. So your interfaces
void merge(int a[],int start,int mid,int end);
void mergesort(int a[],int start,int end);
are exactly the same as
void merge(int* a,int start,int mid,int end);
void mergesort(int* a,int start,int end);
So actually for arrays, you always pass an address to your function.

Answer 3

I don't know why you say:
Here I just passed the value (pass by value) in the funtions without using pointers.
because, when you use
mergesort(int a[],int start,int end)
merge(int a[],int start,int mid,int end)
Then a[] is as same as when you use *a. It is using the pointer way.