What is the difference between char *s[] and char (*s)[]?
c, pointers
Solution
In C, it is better to speak out the declaration. Then, it becomes intuitive. For this, you follow the right-left convention. Here is how it goes:
char *s[5];
How do you speak it? For that you start at the right of the variable name `s` and then go to the left. So you start by saying "s is a/an", On the right, you see a `[]` and you say "s is an array ... ". And then you go to the left and see the `*`, and say "s is an array of pointers." And that is what it is. It si an array of 5 pointers. You can store different pointers in this array.
Now for the other one:
char (*s)[5];
You start the same way. "s is a/an", and then look at the `()`. Anything within the `()` is bound to `s` closer than anything outside. So the `*` is more closely bound with `s` than `[]`. So You now say, "s is a pointer ..." and now you go out of the parenthesis and see the `[]`. So you continue, "s is a pointer to an array". And that is exactly what it is. It is a pointer, which will point to the first element of an array.
Now follow the same logic and try to guess what the following will be:
int (*callme)(int a, int b)
int (*callme[10])(int a, int b)
Hint, the last one can be used to create lookup tables for functions.
Edit:
As mentioned in the comments, there is also a `char` in the beginning. I have never ben able to figure out an easy way of speaking this, but is generally clear from the context. For example, in the first example, the `char` defines the type of the array, while in the second example, it defines pointer. In the exercises I have posted, the `int` defines the type for the return values of the functions. Generally with definitions such as these, there will be exactly one item with the undefined type. And thats how I figure out where the type goes.
Problem
When I read books about C language, the two level pointer bothered me a lot. ``` char s[5][5]; char *s[5]; char (*s)[5]; ``` so what is the difference between them?