Question

I need to take a char [] array and copy it’s value to another, but I fail every time.

It works using this format:

char array[] = { 0x00, 0x00, 0x00 }

However, when I try to do this:

char array[] = char new_array[];

it fails, even though the new_array is just like the original.

Any help would be kindly appreciated.

Thanks

Answer 1

To copy at runtime, the usual C method is to use the strncpy or memcpy functions.

If you want two char arrays initialized to the same constant initializer at compile time, you’re probably stuck with using #define:

#define ARRAY_INIT { 0x00, 0x00, 0x00 }
char array[] = ARRAY_INIT;
char new_array[] = ARRAY_INIT;

Thing is, this is rarely done because there’s usually a better implementation.

EDIT: Okay, so you want to copy arrays at runtime. This is done with memcpy, part of <string.h> (of all places).

If I’m reading you right, you have initial conditions like so:

char array[] = { 0x00, 0x00, 0x00 };
char new_array[] = { 0x01, 0x00, 0xFF };

Then you do something, changing the arrays’ contents, and after it’s done, you want to set array to match new_array. That’s just this:

memcpy(new_array, array, sizeof(array));
/*        ^         ^          ^
          |         |          +--- size in bytes
          |         +-------------- source array
          +-------------------------destination array
 */

The library writers chose to order the arguments with the destination first because that’s the same order as in assignment: destination = source.

There is no language-level built-in means to copy arrays in C, Objective-C, or C++ with primitive arrays like this. C++ encourages people to use std::vector, and Objective-C encourages the use of NSArray.

I’m still not sure of exactly what you want, though.