c qsort string array

Question

    qsort(words, size1, size2, compareWords);

inside compare words:

    int compareWords(const void *ac, const void *bc)

this works:

    char const *a = *(const char **)ac;

these don't (a gets some garbage values):

    char const *a = ac;
    char const *a = (const char *) ac;

what is the rationale?

Also, in some examples I see size2 to be sizeof(char *). Shouldn't this be sizeof(*words)?

words is declared as: char *words[] = {"abc", "pqr", "abcd", "pqsl"};

You must show the declaration of words, since you're asking questions about how to access it. — unwind
– unwind, Commented Aug 23, 2013 at 10:23
You're right, size2 is the size of one element. So, size2 = sizeof(*words) or better size2 = sizeof(char). — Michael M.
– Michael M., Commented Aug 23, 2013 at 12:10
@Michael is sizeof(* words) = sizeof(char)?? In my example words array, sizeof(* words) comes out to be 4 whereas sizeof(char) = 1. I don't think they are the same. — Sushil
– Sushil, Commented Aug 23, 2013 at 20:26
@Sushil, yes my bad... *words is a char* so yeah sizeof(char*) is the same as sizeof(*words) — Michael M.
– Michael M., Commented Aug 24, 2013 at 16:17

Medinoc · Accepted Answer · 2013-08-23 10:28:11Z

3

When qsorting an array of T, your comparison function must convert its const void* pointers to const T*, because T can't be taken by value.

If words is an array of char* or char const *, you have to convert the arguments to char* const * or char const * const * respectively, it's natural when said this way.

answered Aug 23, 2013 at 10:28

Medinoc

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2 Comments

Sushil Over a year ago

hey, thx, but to be precise I wanted to know whats with *(const char **)ac - seems pretty un-intuitive to me... what is wrong in typecasting with (const char*)?

Medinoc Over a year ago

You have an array of char*, and the comparison function always receives pointers to elements in the array. Therefore, pointers to char*. These pointers also happen to be const, so *(const char **)ac is not const-correct, though there are no actual consequences since you don't try to write to this pointer.

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