What's wrong with increment operator in pointer array?

Question

I just found a little confusion while using increment operator in pointer array.

Code 1:

int main(void) {
     char *array[] = {"howdy", "mani"};
     printf("%s", *(++array));
     return 0;
}

While compiling, gcc throws a well known error "lvalue required as increment operand".

But, when I compile the below code it shows no error!!! Why?

Code2:

int main(int argc, char *argv[]) {
     printf("%s",*(++argv));
     return 0;
}

In both cases, I have been incrementing an array of pointer. So, it should be done by this way.

char *array[] = {"howdy","mani"};
char **pointer = array;
printf("%s",*(++pointer));

But, why code2 shows no error?

Very interesting! My guess is that it considers `char *argv[]` equivalent to `char **argv` but not so with a user-defined pointer to array.. — Claudiu, Oct 19 '15 at 00:48
A local variable `char *array[]` is an array of pointers. A function argument `char *argv[]` is actually a `char **argv`. C11 draft standard `6.7.6.3 Function declarators (including prototypes), Section 7 A declaration of a parameter as ‘‘array of type’’ shall be adjusted to ‘‘qualified pointer to type’’[...]`. — EOF, Oct 19 '15 at 00:49
`char *argv[]` passed to `main()` decays to `char **`, not so with your direct use of `*array[]` in `main()` — David C. Rankin, Oct 19 '15 at 00:50

score 5 · Accepted Answer · edited May 23 '17 at 12:14

Arrays cannot be incremented.

In your first code sample you try to increment an array. In the second code sample you try to increment a pointer.

What's tripping you up is that when an array declarator appears in a function parameter list, it actually gets adjusted to be a pointer declarator. (This is different to array-pointer decay). In the second snippet, char *argv[] actually means char **argv.

See this thread for a similar discussion.

What's wrong with increment operator in pointer array?

1 Answers1