Passing an array to a sort function in C language

Question

#include<stdio.h>
#include<conio.h>

float smallest(int arr[],int k,int n);
void sort(int arr[],int n);

void main()
{
 int arr[20],i,n,j,k;
 clrscr();
 printf("\nEnter the number of elements in the array: ");
 scanf("%d",&n);

 printf("\nEnter the elements of the array");
 for(i=0 ; i < n ; i++)
 {
  printf("\n arr[%d] = ",i);
  scanf("%d",&arr[i]);
 }

 sort(arr,n);
 printf("\nThe sorted array is: \n");
 for(i=0 ; i < n ;  i++)
 printf("%d\t",arr[i]);
 getch();
}

int smallest(int arr[],int k,int n)//smallest function
{
 int pos=k,small=arr[k],i;
 for(i=k+1;i<n;i++)
 {
  if(arr[i]<small)
  {
   small=arr[i];
   pos=i;
  }
 }
 return pos;
}


void sort(int arr[],int n)//sorting function
{
 int k,pos,temp;
 for(k=0 ; k < n ; k++)
  {
   pos=smallest(arr,k,n);
   temp=arr[k];
   arr[k]=arr[pos];
   arr[pos]=temp;
  }
}

In the above program the sort function is being called from main but the return type of sort is void and it still returns the sorted array. As after sorting the array the function should return the sorted array back to the calling function to print the sorted array but the program runs perfectly. How is that happening?

Answer 1

When you declare

int arr[20];

you can say "arr is an array of 20 integers". But arr is a pointer to an integer as well, pointing to the first integer in a row of 20. So de-referencing *arr is an integer, the same as arr[0] in fact.

This means when you pass arr to a function you only pass a pointer to that function. The function in this case works on the (copied) pointer. But this very pointer points exactly to the same memory as your original arr declared in main(). And that's the reason why manipulating arr in sort() is in fact manipulating arr in main().

Answer 2

When passing an array as a parameter, this

int smallest(int arr[],int k,int n)

means exactly the same as

int smallest(int *arr,int k,int n)

For example

#include<iostream>
void printArray(int data[])
{
  for(int i = 0, length = sizeof(data); i < length; ++i)
  {
    std::cout << data[i] << ' ';
  }
  std::cout << std::endl;
}

int main()
{
  int data[] = { 5, 7, 8, 9, 1, 2 };
  printArray(data);
  return 0;
}

You will see that only the first 4 elements of the array are printed. The sizeof(data) returns a value of 4! That happens to be the size of the pointer used to pass the array to printArray(). First the array does not get copied. The pointer to the first element of the array is copied

Answer 3

First, there is no connection between any function argument what is, or is not passed using a return statement with an expression according to the function's return type.

While it is true that all parameter passing in C is by value - copy the value to a "local parameter variable" - nothing is assumed about what is to happen at the memory location a pointer is referencing. So, a function can make any changes in the calling environment, even without returning a value.

As to parameters declared as being aType name[]. this is merely syntactic sugar for const aType* name.