Question

I'm having trouble printing a fibonacci sequence using pointers with malloc's to get this output.

now printing with pointers:

0 1 1 2 3 5 8 13 21 34 55 89

Here's my code:

#include <stdio.h>

#include <stdlib.h>

void fib1(int a[]);

void fib2(int* a);

}

int main() {

  int arr[2] = {0,1};

  int *pointer;

  arr[0]=0;

  arr[1]=1;

  fib1(arr);

  

  return 0;

}

void fib1(int a[]){

  printf("printing with arrays:\n");  

  

  for(int i=0; i<6; i++){

    printf("%d %d ", a[0],a[1]);

    a[0] = a[0] + a[1];

    a[1] = a[0] + a[1];

  }

  printf("\n\n");

}

void fib2(int* a){

  printf("printing with pointers:\n");  

  

  for(int i=0; i<6; i++){

    printf("%d %d ", a,a);

    

  }

}

Answer 1

#include <stdio.h>

#include <stdlib.h>

void fib1(int a[]);

void fib2(int* a);

int main() {

int arr[2] = {0,1};

int *pointer;

arr[0]=0;

arr[1]=1;

fib1(arr);
//creating array with size 2 using malloc
pointer=(int *)malloc(2*sizeof(int));
// initiliazing the values 0 and 1
pointer[0]=0;

pointer[1]=1;

fib2(pointer);

return 0;

}

void fib1(int a[]){

printf("printing with arrays:\n");

  

for(int i=0; i<6; i++){

printf("%d %d ", a[0],a[1]);

a[0] = a[0] + a[1];

a[1] = a[0] + a[1];

}

printf("\n\n");

}

void fib2(int* a){

printf("printing with pointers:\n");

  

for(int i=0; i<6; i++){

// so here we need to print the 0 th and 1th element
// so we need to use * operator
// to print 1th element we need to add 1 to the pointer
printf("%d %d ", *(a+0),*(a+1));
a[0] = a[0] + a[1];

a[1] = a[0] +a[1];

  

}

}