Thread: Newbie question: char const **

I'm trying to figure out the third argument in te inotifytools_watch_recursively_with_exclude() method in inotify-tools. It's the third argument and the docs says it's type is "char const ** exclude_list". I don't know what a "char const **" is and it's not in my O'Reily book.

Here the link to the docs: libinotifytools: inotifytools/inotifytools.h File Reference

I was hoping someone could provide a hard coded example. IE, FileA, FileB.

-Thanks,
i

The declaration char const ** is a pointer variable which points to the same location ie a constant pointer, as in

Code:

char const **p;     /* p points to the same location and can't be modified */
const char *x;      /* x is a pointer to a constant char */

Note the position of the const keyword which makes all the difference between the two declarations

I ended up getting the code running with this pretty ugly block of code:

Code:

	char first[100];
	char second[100];
	char third[100];
	char fourth[100];
	strcpy( first, "/c/gitshare/.AppleDouble" );
	strcpy( second, "/c/gitshare/Network Trash Folder" );
	strcpy( third, "/c/gitshare/Temporary Items" );
	strcpy( fourth, "/c/gitshare/.TemporaryItems/" );
	char *strings[] = { first, second, third, fourth, NULL };

Is there a prettier way to do that in C?

What you have is fine though you can initialize the array strings[] using string literals directly:

Code:

char *strings[] = {"first item", "second item", "third item", "fourth item", "fifth item", "\0"};

Thanks.

Originally Posted by itCbitC

What you have is fine though you can initialize the array strings[] using string literals directly:

Code:

char *strings[] = {"first item", "second item", "third item", "fourth item", "fifth item", "\0"};

This is incorrect. The items in your code are constant strings, which you store in non-const char. As soon as it is written to, your program will likely crash. Furthermore, the last "\0" is an empty string, identical to "", except that the former contains two 0-bytes. I expect it should read NULL, although granted, it depends on the called function what it expects.

Originally Posted by Elysia

const char** and char const** should be the same thing. What is left of the * is the type the pointer points to.

Dang! mistook the location of "*" and yep you're right - const char * and char const * are identical; however const char * and char *const are different.

Originally Posted by EVOEx

This is incorrect. The items in your code are constant strings, which you store in non-const char. As soon as it is written to, your program will likely crash. Furthermore, the last "\0" is an empty string, identical to "", except that the former contains two 0-bytes. I expect it should read NULL, although granted, it depends on the called function what it expects.

Last item should be '\0' (copy and paste). The array of pointers to string literals won't let you modify its elements; however it was unspecified what was to be done with 'em.

Originally Posted by Elysia

Yes, those are different. Easy to mistake, so subtle, yet such a big difference.

concur!

Originally Posted by itCbitC

Last item should be '\0' (copy and paste). The array of pointers to string literals won't let you modify its elements; however it was unspecified what was to be done with 'em.

Then why does "strings[0][0] = 'a';" crash in your application? In the line of code you wrote, the string literals can't be modified. So you should store them in a CONST character array, which you didn't.

And what should the last item be? "\0", '\0' or NULL? Arguably the last two are the same though.