Question

I’m having some trouble with lists in C.

I made this struct:

typedef struct str {
char * str;
struct str * prox;
struct str * ant;
} cadena;

Then in the main I start the list with its first pointer to NULL.

cadena * lista = NULL;

Then I made a function where the user can add new elements to the lists (strings in the “str” member) and I manage the pointers to build the list. That seems to work OK. I hope.

But when I want to print a member of one node, it I’m trying to use something like this, and its not working:

void showdata (cadena *lista) {
    int i=0;
    while (lista.str[i] != '\0')
    {
        printf("%c\n",str[i]);
        i++;
    }
}

I’m getting these errors:

from line "while (lista.str[i] != '\0')"
.error: request for member ‘str’ in something not a structure or union 

(why is it expecting a structure? I thought I’m accessing a member there)

from line "printf("%c\n",str[i]);"     
.error: ‘str’ undeclared (first use in this function)

(isn’t it declared with the struct?)

I’m surely doing something really wrong, but I cannot understand which is the right way to access a char member and print it.

Thank you in advance.

Answer 1

It’s a pointer notation problem. C can be confusing in this way.

lista.str[i]

doesn’t work because C doesn’t recognize the “.” as moving to a field. This can be a problem for people like myself who were used to C# or other OOP languages, where you can use the “.” to modify properties of an object, like

control.visibility = "False"

In C, structs are based on pointers, so you need to use the -> like Keith said. You’re pointing to a location in memory, or the “location” in the struct where a certain field is. Think of it that way, and you shouldn’t have too many problems.

C also allows you to use the “.”… but only if you strongly type the reference. For you, that would mean going

(*lista).str[i]

so that your reference to the lista instance is seen as a pointer to the struct.

So, in review, either

lista->str[i]

or

(*lista).str[i]

should work.