I study C language and I have some difficulty to understand pointers and arrays.
In tutorial that I read I have this row:
char* arrP1[] = { "father","mother",NULL };
And my question is what is arrP1?
Is it array of pointers to the static strings:
Or it is pointer to the string array:
I am confused...what is arrP1?
arrP is an array of char * which in this case is an array of size 3, and you've assigned the pointers to c-style strings with initial values of {"father", "mother", NULL}, which themselves are character arrays that are null terminated. So, your first answer is correct.
So, your second answer is correct -- Because... ? Your description suggests to me the first answer.
string(*)[], which doesn't work because C doesn't have a string type.
To find answer for such declerations, you can use cdecl. It'll highly likely answer you.
declare arrP1 as array of pointer to char
However, there is a something that is called as spiral rule. It can also help you to read decleration. For example,
char *str[10]
+-------+
| +-+ |
| ^ | |
char *str[10];
^ ^ | |
| +---+ |
+-----------+
- str is an array of 10 elements
- str is an array of 10, of pointers
- str is an array of 10, of pointers, of type char
char *str[10], that gives str, then [10], then *, then char, so str is an array of ten pointers to char.char **x[10][20][30] looks nothing like a spiral.Not sure if this will help or make things more confusing but arrP1 can be both an array of char* and a char**, like so:
void foo1(char** arr) { cout << arr << endl; }
void foo2(char* arr[]) { cout << arr << endl; }
int main() {
char *arr[] = {"a", "b"};
cout << arr << endl;
foo1(arr);
foo2(arr);
return 0;
}
The interesting thing (that I just found out myself) is that foo2 doesn't create a copy of the array on its stack, it's passed arr directly! All 3 cout print the same address.
char arrp1[]orchar* arrp1, both of which essentially mean the same thing.char* arrp1[]means something else.char *arrp1[]means. Andchar *arrp1andchar arrp1[] = { ... }are semantically very different.char *x = "foo";andchar x[] = "foo";are semantically very different