'Struct' in pure C

Question

I'm working on a little project in pure C. I use gcc compiler (and thats the idea). I have a struct as follows:

struct STACK{ char var; struct STACK* next; struct STACK* prev; };

and little lower in my function f():

STACK head; head->var='a'; printf("%c", head->var);

Now, during compilation I get the error:

: In function ‘f’: :13:3: error: unknown type name ‘STACK’ :14:7: error: invalid type argument of ‘->’ (have ‘int’) :15:20: error: invalid type argument of ‘->’ (have ‘int’)

Could someone explain me what is wrong ? When I was working in C++ and g++ everything seemed fine.

================= EDIT:

OK, that worked, but now I have to dynamically create structures of type STRUCT and only keep pionters to them. Is it possible in C ?

sh03 · Accepted Answer · 2014-02-02 01:31:57Z

You need to add the struct keyword before the struct name STACK:

struct STACK head;

instead of:

STACK head;

Also you are using -> which is usually applied to pointers. Just use the . instead:

head.var

Remember that:

x->y

can be translated to:

(*x).y

Finally if you define a struct as:

typedef struct { char var; struct STACK* next; struct STACK* prev; } STACK;

you can declare the variable as:

STACK head;

danh · Accepted Answer · 2014-02-02 01:20:12Z

1

The definition is okay, but to declare a variable, try:

struct STACK head;

And to refer to it's members,

head.var = 'a';

answered Feb 2, 2014 at 1:20

danh

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2 Comments

maciek Over a year ago

Thank you very much, that worked. Can you please tell me why arrow notation does not ?

danh Over a year ago

The arrow is used when head contains a pointer to a structure. When the variable is the struct itself, we refer to the members with dot.

mathematician1975 · Accepted Answer · 2014-02-02 01:20:30Z

As you have not provided a typedef for your struct you need to declare as

struct STACK head;

also head is not a pointer so the use of -> should be replaced with . when you accees its members

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