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
– Claudiu, Commented Oct 19, 2015 at 0:48
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’’[...]. — user3185968
– user3185968, Commented Oct 19, 2015 at 0:49
char *argv[] passed to main() decays to char **, not so with your direct use of *array[] in main() — David C. Rankin
– David C. Rankin, Commented Oct 19, 2015 at 0:50

Community · Accepted Answer · 2017-05-23 12:14:33Z

5

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.

edited May 23, 2017 at 12:14

CommunityBot

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answered Oct 19, 2015 at 1:06

M.M

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