Why we cannot create an array with a constant in c

Question

Why the below piece is a valid C++ code but an - invalid C code?

int main(){
    unsigned int const size_of_list = 20;
    const char* list_of_words[size_of_list] = {"Some", "Array"};

    for (unsigned int writing_index = 0; writing_index < size_of_list; writing_index ++)
        ;
        return 0;
}

So your question is why does the `c` compiler reject this code but the `c++` compiler accept? — drescherjm, Dec 28 '21 at 17:55
The error message is pretty clear. You're allowed to declare variable-length arrays, but you can't use an initializer with them. — Barmar, Dec 28 '21 at 17:56
This question may need more focus if you are really asking the two questions "Why is this valid C++?" and "Why is this invalid C?" Otherwise there could be a near-infinite number of questions like this. — Drew Dormann, Dec 28 '21 at 17:58
@DrewDormann The title seems to indicate that the emphasis is on C. — Barmar, Dec 28 '21 at 18:10
@FrançoisAndrieux I reopened it, but there's almost certainly a duplicate for that question as well. — Barmar, Dec 28 '21 at 18:11
Just for general interest: gcc will compile the posted code, albeit with a warning of "variable length array folded to constant array as an extension [-Wgnu-folding-content]" if you specify -pedantic on the compile line. — Jeremy Friesner, Dec 28 '21 at 19:48

score 3 · Accepted Answer · edited Dec 28 '21 at 21:08

In C:
20 is a constant.
unsigned int const size_of_list is not a constant.

Title: "Why we cannot create an array with a constant in c" does not apply to this code.

const char* list_of_words[size_of_list] = {"Some", "Array"}; // Bad

An issue here (and the error message) is why a VLA cannot be initialized. That is answered here.

With a constant, array initialization works fine.

const char* list_of_words[20] = {"Some", "Array"};  // Good

Another issue is mistaking that const makes an object a constant. It does not.

Alternative C code

int main(){
    // unsigned int const size_of_list = 20;
    #define size_of_list 20u

    const char* list_of_words[size_of_list] = {"Some", "Array"};
    for (unsigned int writing_index = 0; writing_index < size_of_list; writing_index ++)
        ;
    return 0;
}

If you can specify the size in the array itself, then you can get it's size with the sizeof operator. This may be better suited for having the compiler count up the size instead manual counting. When not using C99 VLAs, sizeof also yields a compile-time constant.

#include <stddef.h> /* size_t */
int main(void) {
    const char* list_of_words[] = {"Some", "Array"};
    const char list_of_char[sizeof list_of_words
        / sizeof *list_of_words] = {'S','A'};
    const size_t size_of_list
        = sizeof list_of_words / sizeof *list_of_words;
    for (size_t writing_index = 0;
        writing_index < size_of_list; writing_index ++);
    return 0;
}

Thanks, @chux. What is the significance of const then? Could please help me by pointing to some good resources on this? — thumala manish, Dec 28 '21 at 20:20
`const` directs code to not subsequently change `size_of_list`. Aside from the C spec, no suggested resources. — chux - Reinstate Monica, Dec 28 '21 at 20:25

Why we cannot create an array with a constant in c

1 Answers1