Dereferencing array pointers

Question

I'm trying to dereference this pointer to a new array (ARR) , but from this example below when i do the derefencing *issub it only "carries" the first letter. How can i fix this situation and have ARR(subscount) be the word i want?

#include <iostream>
  int main(){
  char inp[3]={'O','I','L'};
  int lll=3;
  char ARR[3];
  int subscount=0;

  char * issub= new char[lll];
  for(int i=0;i<lll;i++){
    issub[i]=inp[i];
  }
  ARR[subscount]=*issub;
 }

Your code is bad because inp is being accessed with out-of-bounds index! — Poriferous, Apr 10 '16 at 20:25
What exactly do you want to do? Fill every cell of `ARR` so it will be the same with `inp`, **or** have a pointer in `ARR[subscount]` will will point to the word? — gsamaras, Apr 10 '16 at 20:25
@gsamaras But if i print issub before the ARR it prints the whole word — J. Barbosa, Apr 10 '16 at 20:31
Dynamic arrays are a very odd language feature. This is the only time that an object exists whose type is not the type of any expression in the language. You could reinterpret cast to `char(*)[111]` if you wanted to, but that's really fighting the language (and doesn't work if you have dynamic size). — Kerrek SB, Apr 10 '16 at 20:34
Are you looking for some way to copy a string? If so, `strcpy` is what you want. — David Schwartz, Apr 10 '16 at 21:09
@J.Barbosa that causes undefined behaviour; there is no null termination on `issub` — M.M, Apr 10 '16 at 21:13
@DavidSchwartz `strcpy` does not work on char arrays that do not contain strings (as in this question) — M.M, Apr 10 '16 at 21:13

score 1 · Accepted Answer · edited May 23 '17 at 12:08

I feel that you are confused, so I made an example for you:

#include <iostream>

int main() {
  char inp[4] = {'O','I','L', '\0'};
  int lll = 4;
  // initialize elements of 'ARR' to 0 for safety
  char ARR[4] = {0};
  int subscount = 0;

  // dynamic allocation, DO NOT forget to de-allocate
  char* issub = new char[lll];
  // copy every element of 'inp' to 'issub'
  for(int i=0;i<lll;i++) {
    issub[i]=inp[i];
  }
  // this will copy where '*issub' points to,
  // that is the 'issub[0]', to 'ARR[subscount]'
  ARR[subscount] = *issub;
  std::cout << ARR << "\n"; // prints: O
  // you can use a for loop as before to copy the contents,
  // from 'inp', 'issub' to 'ARR'

  // However, we will do something different here,
  // so I am de-allocating 'issub'
  delete [] issub;

  // We will use an array of pointers, with size 2,
  // thus it will have two pointers in total.
  char* ptr_arr[2];

  // Assign the first pointer to 'inp'
  ptr_arr[0] = &(inp[0]);

  std::cout << ptr_arr[0] << "\n"; // prints OIL

  // we don't use the second pointer,
  // set it to NULL
  ptr_arr[1] = NULL;
  return 0;
}

Hope that helps (but it really reminds me of C, rather than C++, where std::string should be used).

Updated with a null terminated string; What is a null-terminated string?

_{Thanks @M.M}

I think i got it down,but shouldn't `inp[0]` be just the first character? — J. Barbosa, Apr 10 '16 at 20:55

Dereferencing array pointers

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