C Programming - Pointers - Discussion

Cannot convert an integer to a character pointer.

So,to get the desired output the revised code should be p=(char*)arr in place of p=arr and p=(char*)(int*)(p+1) in place of p=(int*)(p+1) .

Simply p=p+1 works. No need to convert integer pointer then character pointer.

Just, (p+1) is NULL. (p+2) is NULL. (p+2) is NULL. (p+4) is NULL. etc
Otherwise, p=(int*)(p+1) is NULL or ZERO.

It will give a warning instead of error hence p=arr is illegal but this code will compile and give output.

But my question is how to know that these integers will be stored in little-endian firm or big-endian?

Please, anyone, help me.

All droughts are cleared, Thanks for explaining @Puru.

@All.

According to me; the coding part is;


#include<stdio.h>

int main()
{
// create and array hear with name arr

int arr[3] = {2, 3, 4};

// assume array is created at memory location 1000
// so 2 is store at location 1000 and
// 3 on 1004 and
// 4 is store at 1008......

// created one char pointer *p
char *p;
//p is pointing to the 1000
p = arr;

p = (char*)((int*)(p));
printf("%d, ", *p);
p = (int*)(p+1); // when you increase hear p+1 then 1000 becomes 1001 because p is character pointer
printf("%d", *p);
return 0;
}

Well done @Mayur.