Array of char pointers

Question

I am looking at some code I did not write and wanted help to understand an element of it. The code stores character arrays, creates pointers to these arrays (assigning the pointers the arrays addresses). It looks like it then creates an array to store these characters pointers addresses and I just wanted some clarification on what I am looking at exactly. I also am confused about the use of the double (**) when creating the array.

I have included a stripped down and simplified example of what I am looking at below.

char eLangAr[20] = "English";
char fLangAr[20] = "French";
char gLangAr[20] = "German";

char* eLangPtr = eLangAr;
char* fLangPtr = fLangAr;
char* gLangPtr = gLangAr;


char **langStrings [3]=
{
    &eLangPtr,
    &fLangPtr,
    &gLangPtr
};

When using the array they pass it as an argument to a function.

menu (*langStrings[0]);

So the idea is that the character array value "English" is passed to the function but I'm having trouble seeing how. They pass the menu function a copy of the value stored in the langStrings function at location 0, which would be the address of eLandPtr? If someone could explain the process in English so I could get my head around it that would be great. It might be just because it has been a long day but its just not straight in my head at all.

Answer 1

You are correct that langStrings contains pointers to pointers of array of characters. So each langString[i] points to a pointer. That pointer points to an array. That array contains the name of a language.

As others point out, it looks a bit clumsy. I'll elaborate:

char eLangAr[20] = "English"; is an array of 20 chars and the name "English" is copied to it. I don't expect that variable eLangAr will ever contain something else than this language name, so there is no need to use an array; a constant would be sufficient.

char **langStrings [3]= ... Here, it would be sufficient to have just one indirection (one *) as there seems no need to ever have the pointer point to anything else (randomly shuffle languages?).

in conclusion, just having the following should be sufficient:

const char *langStrings [3]=
{
    "English",
    "French",
    "German"
};

(Note the const as the strings are now read-only constants/literals.)

---

What the given code could be useful for is when these language names must be spelled differently in different languages. So "English", "French", "German" become "Engels", "Frans", "Duits". However, then there still is one level of indirection too many and the following would be sufficient:

char *langStrings [3]=
{
    aLangArr,
    fLangAr,
    gLangAr
};