why does my array of strings assignment not work with pointers?-CodePudding

I have a struct that stores 5 char * pointers

struct x {

       char *arr[5];
};

I allocated memory for the struct using malloc()

struct x *str = malloc(sizeof(struct x));

However when i try to initialize arr with a value (namely a read only string-literal) it gives me an error

error: expected identifier before ‘(’ token
   13 |     (*str).(*(arr 0)) = "hello";
                   ^

here is my initialization

(*str).(*(arr 0)) = "hello";

i know i could do it like this

str->arr[0] = "hello"

but i would like to understand how array of strings work, so i have used pointers and firstly dereferenced str-> and changed it to (*str).

Also since arr[0] works in the initialization str->arr[0] = "hello" and i know that arr[0] is equivalent to *(arr i) where i is a array cell, i though that this would work in (*str).(*(arr 0)) = "hello"; but it does not.

Why is this and how are arrays of strings actually working behind the scenes?

CodePudding user response：

The structure has the data member arr.

struct x {

       char *arr[5];
};

So you have to write accessing the data member using the operator . or -> for example like

*( (*str).arr   0) = "hello";

that is the same as

*(*str).arr = "hello";

*str->arr = "hello";

str->arr[0] = "hello";

That is member access operators are defined like

postfix-expression . identifier 
postfix-expression -> identifier

while you are trying to use an expression that is not an identifier.

Using your approach to access ant element of the array you can write

*( (*str).arr   i) = "hello";

*( str->arr   i) = "hello";

That is at first you need to get the array designator like

( *str ).arr

str->arr

and then you can use the pointer arithmetic with the obtained array.