Traverse an array using malloc

Question

In the course of some questions here in the SO I saw this example and I was left with doubts.

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

void main()
{
clrscr();
int *ptr,*temp;
int i;

ptr = (int *)malloc(4*sizeof(int));
temp = ptr;  

for(i=0;i < 4;i++)
     {
     printf("Enter the Number %d : ",i);
     scanf("%d",ptr);
     ptr++;         
     }

ptr = temp;
   for(i=0;i < 4;i++)
   {
   printf("\nNumber(%d) : %d",i,*ptr);
   ptr++;
   }

getch();
}

What is really happening in this example?

Why do you wear ptr++ and not ptr[x]?

malloc does not always allocate continuous memory?

Answer 1

What is really happening in this example?

For all intents and purposes you are accessing a array, albeit technically the definition is a little different. In C the array is always accessed via pointers.

Why is ptr++ used and not ptr[x]?

Because who made the code wanted it so. Both work. It is common to use pointer notation when using pointer and notation array when using array. ptr[x] is only syntactic sugar for *(ptr + x). The pointer shape is changing the pointing location so it needed to create a temporary variable to hold the original start value of the array. generally if you need a temporary variable the ideal would be the notation of array.

It is possible to use this formula ptr[x]?

I don’t know what that means.

malloc does not always allocate continuous memory?

Yes, one has nothing to do with the other.

This code could be written like this:

#include<stdio.h>
#include<stdlib.h>

void main() {
    int *ptr = malloc(4 * sizeof(int));
    int *temp = ptr;  
    for (int i = 0; i < 4; i++) {
         printf("\nEnter the Number %d : ", i);
         scanf("%d", temp++);
    }
    for (int i = 0; i < 4; i++) printf("\nNumber(%d) : %d", i, *ptr++);
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.