How to declare a variable as char* const*?

Question

I have a bluez header file get_opt.h where argv is an argument:

 extern int getopt_long (int ___argc, char *__getopt_argv_const *___argv,.....

which requires char* const* for argv. Because the unit is called from a different file I was trying to emulate argv but whatever permutation of declaring char, const and * is used I get unqualified-id or the compiler moans about not converting to char* const*. I can get const char * easily, of course. char const * argv = "0"; compiles OK in Netbeans with C++14 but

char * const * argv = "0";

produces

"error: cannot convert 'const char *' to 'char * const *'" in initialisation (sic).

How do you declare char* const* or am I mixing C with C++ so I'm up the wrong tree?

You can declare a variable as `char* const*` by declaring it as `char* const*`. It is hard to determine why this did not work for your without a [mre] and the error message (copy-pasted, not interpreted). — JaMiT, Oct 08 '21 at 18:05
Pick the example that you feel was closest to correct, and the first error reported. Maybe use a horizontal rule (`---` on a line by itself) and provide a second example (and its first error message) as a footnote, if this does not become distractingly long. — JaMiT, Oct 08 '21 at 21:54
@JaMiT Understand about no reproducible example but there were/are so many. char * const * did not work in this situation. char const * argv = "0"; compiles OK in Netbeans with C++14 but char * const * produces error: cannot convert 'const char *' to 'char * const *' in initialisation (sic) — Hillybilly, Oct 08 '21 at 22:03
Good, that's an example. Please add it to your question. Also, you might want to notice how you can remove the call to `getopt_long()` from that example while retaining the error (it's motivation for your declaration, but not the source of your issue). If you think about your presentation enough, you might realize that your question is not how to *declare* a `char* const*` but how to *initialize* it. — JaMiT, Oct 08 '21 at 22:14
Due to arrays decaying to a pointer, you could declare an array of pointers (`char* const argv[]`) instead of a pointer to pointer (`char* const * argv`). See _The 1D array of pointers way_ portion of [A: 2D Pointer initialization](https://stackoverflow.com/a/6465638) for how to initialize such a beast. *Beware:* given your setup, the "2D array way" from that answer is not an option. *Reminder:* The last pointer in the argument vector is null; `argv[argc] == nullptr` when those values are provided by the system to `main()`. — JaMiT, Oct 09 '21 at 15:36

score 2 · Answer 1 · answered Oct 08 '21 at 17:50

2

Let's review the pointer declarations:

char *             -- mutable pointer to mutable data.  
char const *       -- mutable pointer to constant data.  
char       * const -- constant pointer to mutable data.  
char const * const -- constant pointer to constant data.

Read pointer declarations from right to left.

Looking at the descriptions, which kind of pointer do you want?

answered Oct 08 '21 at 17:50

Thomas Matthews

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OP thinks that he has to change the type of main function parameters to match the function one :) – 0___________ Oct 08 '21 at 17:52
@Thomas Saw that before. Clear and helpful thank you but it didn't work in this situation. – Hillybilly Oct 08 '21 at 21:50

0___________ · Answer 2 · 2021-10-08T18:02:37.507

Because void main(void) is in a different file I was trying to emulate argv

void main(void) is not the correct main function declaration. The main function has to have int return value.

Simply change the void main(void) to int main(int argc, char **argv) in that "other" file.

It does not require char const * to be passed to. Do not change the type of the main function arguments.

int getopt_long (int ___argc, char * const *___argv);

int main(int argc, char **argv) 
{
    printf("%d\n", getopt_long(argc, argv));
}

Both language compile fine: https://godbolt.org/z/5To5T931j

score 1 · Accepted Answer · edited Oct 10 '21 at 12:11

Credit @JaMiT above

[I don't know how to accept a commment]

Also, you might want to notice how you can remove the call to getopt_long() from that example while retaining the error (it's motivation for your declaration, but not the source of your issue). If you think about your presentation enough, you might realize that your question is not how to declare a char* const* but how to initialize it. Due to arrays decaying to a pointer, you could declare an array of pointers (char* const argv[]) instead of a pointer to pointer (char* const * argv). See The 1D array of pointers way portion of A: 2D Pointer initialization for how to initialize such a beast. Beware: given your setup, the "2D array way" from that answer is not an option. Reminder: The last pointer in the argument vector is null; argv[argc] == nullptr when those values are provided by the system to main().

How to declare a variable as char* const*?

3 Answers3