-> Arrow operator with Arrays?

Question

Can you use the arrow operator with an array of type struct. For example:

struct {
char *foo;
} fooArray[10];

fooArray[0]->foo = ...

Answer 1

Not in your case because you are accessing a struct (not a pointer to a struct) when you use

fooArray[0]

The compiler would let you know about that problem.

Answer 2

No, you can't. Has to be a pointer. E.g. this is OK syntax-wise (but it will segfault unless fooArray is assigned):

int main ()
{
struct {
char *foo;
}* fooArray[10];

fooArray[0]->foo = "hallo";
}

Answer 3

No, you have to use fooArray[0].foo.

fooArray is a pointer that happens to point to the first element of an array. Using the [] is dereferencing that pointer at the given element so once applied it's no longer a pointer and the -> operator cannot be applied since that operator does both dereferencing and accessing a struct member.

Answer 4

you can use array operator(for accessing data members) only on pointers, fooArray[0] is not a pointer to struct, If you want to use it you take address of it and use it like

(&fooArray[0])->foo="hello";

Or declare pointer/array of pointers to struct and allocate memory to it ..like

fooArray* a=new fooArray();
a->foo="hello";

Answer 5

Remember the value of fooArray by itself is a pointer. You can dereference the pointer, add to it, ...

fooArray->foo; /* same as fooArray[0].foo */
(fooArray+3)->foo; /* same as fooArray[3].foo */

Answer 6

Just remember that a->b is a shorthand for (*a).b. Keeping this in mind should help you tackle any situation. I.e, in your case fooArray[0]->foo means (*fooarray[0]).foo, but fooarray[0] isn't a pointer (it's a struct), so you can not dereference it. Hence it won't work.