Pointer arithmetic in array of strings

Question

I am reading a linux programming with c language book and it has the following code.

        int main(int argc, char *argv[])
    {
        int opt;
        while ((opt = getopt(argc, argv, "iv")) != -1) {
            switch (opt) {
            case 'i':
                opt_ignorecase = 1;
                break;
            case 'v':
                opt_invert = 1;
                break;
            case '?':
                fprintf(stderr, "Usage: %s [-iv] [<file>...]\n", argv[0]);
                exit(1);
            }
        }
        argc -= optind;
        argv += optind;
}

I don't understand that why argv += optind works. When i tried at another program, which is the following:

int main(int argc, char* argv[])
{
    
    char* test[] = {"Hello", "World", "I"};
    argv += 1;
    test += 1;
    printf("%s", test[0]);
    return 0;
}

the gcc compiler shows error

a.c:8:9: error: assignment to expression with array type
8 |    test += 1;

the += on argv works but doesn't work with test. I don't understand which part I did wrong..

char* argv[] as a function parameter is a different way to write char** argv. So, it is a pointer to a pointer, test is an array of pointer — mch
– mch, Commented Nov 17, 2020 at 14:35

Eric Postpischil · Accepted Answer · 2020-11-17 14:42:34Z

3

When an array declaration appears in a function parameter list, it is automatically adjusted to declare a pointer. So int main(int argc, char *argv[]) actually declares argv to have type char **.

Since argv is a pointer, we can increment it with argv += optind;. argv initially points to the first of several char *, and argv += optind; changes which of those char * it points to.

In contrast, the declaration char* test[] = {"Hello", "World", "I"}; defines test to be an array of char *. Since it is an array, it cannot be moved; test += 1 attempts to change where the array is, and that is not allowed.

After defining test, you could define a new pointer, char **p = test;, and then you could adjust p with p += 1;. (Note that in char **p = test;, test is automatically converted to a pointer to its first element, so this is equivalent to char **p = &test[0];.)

(Note: Grammatically, the * in char * test [] binds with test [], not char, so char* test[] is misleading. A declaration such as char* a, b; would not declare both a and b to be char*. It declares a to be char * and b to be char, so char *a, b; is a better picture of the meaning.)

edited Nov 17, 2020 at 14:42

answered Nov 17, 2020 at 14:37

Eric Postpischil

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2 Comments

nicklowkc Over a year ago

Hi eric, thank you for your explanation. After trying to do something like char *hello[] = {"Hello", "I", "am", "Low"}; char ** test = hello; it works. By the way, a little out of topic, whats the downside if i do something like above?

Eric Postpischil Over a year ago

@nicklowkc: There is not much downside by itself, and I would not worry about it until you are further along in learning C and programming. One downside is that form requires two identifiers, hello and test, whereas it can be done with one, char **test = (char *[]) {"Hello", "I", "am", "Low"};. But there are other considerations in the context of a program, such as whether the array should be static or automatic, whether you need to reset to the beginning, whether the same strings are used elsewhere, and so on.

John Bode · Accepted Answer · 2020-11-17 14:40:09Z

1

argv doesn't have array type, despite how the declaration is written; its actual type is char **.

In a function parameter list, any declaration of the form T a[N] or T a[] is adjusted to T *a - all three declare a as a pointer, not an array. This is only true for function parameter declarations, not regular declarations.

answered Nov 17, 2020 at 14:40

John Bode

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