[Courses] Re: pointers, array's, and sizeof()

[Laurel Fan]

On Sat, Feb 02, 2002 at 09:04:24PM -0800, Dreaming Kat wrote:
> The last exception, the name of the array being a string literal
> initializer for a character array, means that you've said something
> like:
> 
> char oldArray[ 10 ];
> /* put stuff in oldArray */
> char newArray[] = oldArray;
> 
> this works exactly like you'd expect, the contents of oldArray are
> copied into newArray.  I'll include this in my little sample program
> tomorrow too.

Are you sure?  I don't think you can cause an array copy with an
assignment, even at initialization.

With this code:

  char oldArray[] = "abc";
  char newArray[] = oldArray;

I get a compile error. (array initialized from non-constant array
expression)

Even with this:

  const char oldArray[] = "abc";
  char newArray[] = oldArray;

I get a different error. (invalid initializer)

AFAIK, the only way to intialize an array is with bracket notation,
eg:

  char oldArray[] = {'a', 'b', 'c', '\0'};

or, for arrays of char, a string literal:

  char oldArray[] = "abc";

-- 
laurel at sdf.lonestar.org
SDF Public Access UNIX System - http://sdf.lonestar.org