Converting an array of int to an array of char and back via memcpy in c++

Question

I want to convert an array of int to an array of char and then back again, for serialization purposes. It does not need to work cross-platform.

I came up with

  //sample int array 
  int arr[] = {1,2,3,4,100};
  char char_arr[sizeof(arr)];
  //int arr to char arr
  memcpy(char_arr, arr, sizeof(arr));

  int newarr[sizeof(char_arr)/sizeof(int)];
  //char arr back to int arr
  memcpy(newarr, char_arr, sizeof(char_arr)/sizeof(int));

This does not seem to work however (newarr contains values different from arr). Any suggestions?

While you get the number of entries right in newarr the size when you copy to it is still sizeof(char_arr), i.e. do not divide by sizeof(int). — Some programmer dude
– Some programmer dude, Commented Dec 9, 2011 at 11:42
You don't need to divide by sizeof(int) in the call to memcpy. — n. m. could be an AI
– n. m. could be an AI, Commented Dec 9, 2011 at 11:45
Think again whether you really need this. Perhaps a reinterpret_cast<const char *>(arr) will suffice for your application? — Kerrek SB
– Kerrek SB, Commented Dec 9, 2011 at 13:14

Oliver Charlesworth · Accepted Answer · 2011-12-09 11:41:31Z

6

The third argument to memcpy is number of bytes^*, not number of elements. So your second memcpy call is incorrect.

_{* Number of chars, technically.}

answered Dec 9, 2011 at 11:41

Oliver Charlesworth

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Comments

Werner Henze · Accepted Answer · 2020-05-10 19:14:58Z

0

memcpy copies memory bytes, not array elements and thus does no array element conversion for you. If you want to convert an array of chars to an array of ints you might try std::copy which copies containers element by element.

int newarr[sizeof(char_arr)/sizeof(char_arr[0])]; 
std::copy(&char_arr[0], &char_arr[sizeof char_arr/sizeof char_arr[0]], newarr);

edited May 10, 2020 at 19:14

answered Dec 9, 2011 at 13:12

Werner Henze

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