Append item to a list ( C macro )

[Schmidt]

I have my own declared types and I have to operate with lists of items. I wish to write a one macro, instead of using a function for each type. It's intended to be like this:

#define ____append_to_list(head, item, type) \
type* ptr; \
for ( ptr = head; ptr->next != NULL; ptr = ptr->next ) \
{} \
item->prev = ptr; \
ptr->next = item

but when I call the macro

____append_to_list(list_head, new_item, Item);
/* list_head and new_item is a pointers of a 'Item' type */

and try to compile it, I get gcc error

error: expected expression before 'Item'

how can I use it correctly?...

[artu]

I tested macro in my machine. There is no problem with macro nor get compilation error, but probably your structure declaration is erroneous. Here is my code:

typedef struct list {
struct list* prev;
struct list* next;
} list;

int main(int argc, char **argv) {
list* head = (list *) malloc(sizeof(list));
list* item = (list *) malloc(sizeof(list));
____append_to_list(head, item, list);
return 0;
}

[Schmidt]

I tested macro in my machine. There is no problem with macro nor get compilation error, but probably your structure declaration is erroneous. Here is my code:

hmm, yes, my definition differs from yours, here it is:

typedef struct _Item {
/* ... */
struct _Item *prev;
struct _Item *next;
} Item;

but in other parts of code using Item type it's not a problem. Maybe it's because of compiler options?... I use GCC version 6.8 with no additional compiling or preprocessing options. I'll try to delete the underline sigil first, of course...but still interested in this phenomenon

[artu]

its really interesting. I got no errors with underline or without. its also should not be like that logically.
I compiled as : gcc test.c -o test
my gcc version: 4.4.1 (I think oldie one )

[Schmidt]

))
it seems that my head just has "buffer overflow" I tried to call ____list_append() but macro was defined as ____append_to_list() ... Oh, just words playing, but its price appeared high )
there is another problem too - it's because of

...
type* ptr; \
...

when I use the macro once, that's no problem. But if i call it two or more times in the same namespace (eg. in a function), compiler says that the 'ptr' is redeclared. How can I fix it?

[Schmidt]

I suppose I should not declare pointer inside of macro - it seems that I need just declare a static pointer for each data type and just pass them to my macro...

[sixdrift]

Enclose the body of the macro in {} to make the "ptr" variable be local to that block. This allows you to have that variable defined multiple times locally, so you can have multiple calls to the macro. Like this:

#define ____append_to_list(head, item, type) \
{ \
    type* ptr; \
    for ( ptr = head; ptr->next != NULL; ptr = ptr->next ) \
    {} \
    item->prev = ptr; \
    ptr->next = item; \
}

[Rubberman]

Enclose the body of the macro in {} to make the "ptr" variable be local to that block. This allows you to have that variable defined multiple times locally, so you can have multiple calls to the macro. Like this:

#define ____append_to_list(head, item, type) \
{ \
    type* ptr; \
    for ( ptr = head; ptr->next != NULL; ptr = ptr->next ) \
    {} \
    item->prev = ptr; \
    ptr->next = item; \
}

Just what I was going to say. Good find sixdrift, and welcome to the forums!

[artu]

Enclose the body of the macro in {} to make the "ptr" variable be local to that block. This allows you to have that variable defined multiple times locally, so you can have multiple calls to the macro. Like this:

#define ____append_to_list(head, item, type) \
{ \
    type* ptr; \
    for ( ptr = head; ptr->next != NULL; ptr = ptr->next ) \
    {} \
    item->prev = ptr; \
    ptr->next = item; \
}

Very well sixdrift!

[Schmidt]

Yeah, it works, thanks!