C array pointer arithmetic

Question

I am trying to rewrite my code that takes the user input array, goes to the function and adds zeros between each number and saves it to array 2. My source code works just fine but I am having trouble trying to make it so that it uses pointer arithmetic just for the function to visit each array element, it cannot be sub scripting. What can you tell me about my code or suggestions on how to do this?

Source code:

#include <stdio.h>

void insert0(int n, int a1[], int a2[]);

int main(void) {

  int i;     
  int n;

  printf("Please enter the length of the input array: ");   
    scanf("%d", &n);

  int a[n];   
  int b[2*n];  

  printf("Enter %d numbers for the array: ", n);   
    for (i = 0; i < n; i++){     
      scanf("%d", &a[i]);
    }

  insert0(n, a,  b);

  printf("Output array:");   
    for (i = 0; i < 2*n; i++){
      printf(" %d", b[i]);   
        printf("\n");
    }
    return 0; 
}

void insert0(int n, int a[], int b[]) {

  int i, j = 0; 

  for(i = 0; i < n; i++, j+=2){    
    b[j]= a[i];    
      b[j+1] = 0; 
  }
}

My arithmetic:

   #include <stdio.h>

    void insert0(int n, int *a1, int *a2);

    int main(void) {

      int i;     
      int n;

      printf("Please enter the length of the input array: ");   
        scanf("%d", &n);

      int a1[n];   
      int a2[2*n];  

      printf("Enter %d numbers for the array: ", n);   
        for (i = 0; i < n; i++){     
          scanf("%d", &a2[i]);
        }

//not sure if this is how you call it, I've seen it called with arr
      insert0(n, a1, a2); 

      printf("Output array:");   
        for (i = 0; i < 2*n; i++){
          printf(" %d", a2[i]);   
            printf("\n");
        }
        return 0; 
    }

    void insert0(int n, int *a1, int *a2) {

      int *p;
      int j = 0;

        // I think I translated this properly     
        for(p = a1; p < a1+n; p++, j+=2){
          a2+j = a1+p;  
          //unsure how to get a2[j+1] to still work here with pointer  
            a2(j+1) = 0; 
      }
    }

Answer 1

This is a reply which is not intended to be a demo in efficiency of using pointer arithmetic, but rather to show how array indexing (subscripting) translates into pointer arithmetic, applied to your code. Please, take a look at this SO post containing valuable information from the C standard.

Your code also has few logical bugs, reading into and writing from wrong arrays, etc… You can easily find them yourself. Here's a working fix:

#include <stdio.h>

void insert0(int n, int *a1, int *a2) {
  int p;    //just a normal int for counter
  int j = 0;

    for(p = 0; p < n; p++, j+=2){
      *(a2+j) = *(a1+p);   //values pointed at are getting assigned
                           //equivalent to a2[j] = a1[p]
      if(p < n - 1 )       //we insert only between numbers
        *(a2 + j + 1) = 0; //pointer pointing at memory into which 0 is copied incremented by 1
    }
 }


int main(void) {
  int i;     
  int n;

  printf("Please enter the length of the input array: ");   
    scanf("%d", &n);

  int a1[n];   
  int a2[2*n];  //one element too long, should be a2[n+n-1];

  printf("Enter %d numbers for the array: ", n);   //reading into smaller array
    for (i = 0; i < n; i++){     
      scanf("%d", &a1[i]);
    }

  insert0 (n, a1, a2); //this is OK

  printf("Output array: ");   
  for (i = 0; i < 2*n - 1; i++){
     printf(" %d", a2[i]);              //printing from the bigger, zero-padded array
  }
  printf("\n");
  return 0; 
}