why C has two ways to declare structs?

angustia · 04-14-2006, 08:37 PM

hi, this is my doubt:

there are two ways:

struct { int a,b; } mystruct_s;

and

typedef struct { int a, b; } mystruct_t;

even there's a mixture:

typedef struct mystruct_s { int a,b; } mystruct_t;

is this something about early C standards or it has a purpose?

(the only thing i can guess is that it's so to declare the struct type in the header and the struct details in the .c file but it keeps looking weird...)

thanx.

taylor_venable · 04-14-2006, 10:36 PM

There are a lot of different ways to create a structure in C, and they're all actually a little bit different from one-another.

The basic way is to do something like this:

Code:

struct point {
    int x;
    int y;
};
struct point pt = {0, 0};

This creates pt as a struct of type "point", and sets the x and y members to zero.

Another way is using an anonymous struct:

Code:

struct {
    int x;
    int y;
} pt;
pt.x = 0;
pt.y = 0;

This method creates pt out of an anonymous struct, defined inside the pt declaration. In other words, it cuts out "struct point" part.

A more popular way uses typedef to create a new type:

Code:

typedef struct {
    int x;
    int y;
} point_t;
point_t pt;
pt.x = 0;
pt.y = 0;

This assigns the anonymous struct to the type point_t. By doing this you can treat point_t just like you would any other type to declare new point structs.

The problem with this last method is that members of the struct you are defining can't be of that type, because that type's definition isn't done yet. In other words, this is illegal:

Code:

typedef struct {
    node left;
    node right;
    int value;
} node;

Because node is used before it's defined. The solution, then is to forego the anonymous struct, and use a named one instead. So the following will suffice, for example for a binary tree:

Code:

typedef struct node {
    struct node left;
    struct node right;
    int value;
} node_t;
node_t parent_node;
node_t left_node;
node_t right_node;
parent_node.left = left_node;
parent_node.right = right_node;

And that's how you do trees in C.

addy86 · 04-15-2006, 01:56 AM

Quote:

Originally Posted by taylor_venable

Code:

typedef struct node {
    struct node left;
    struct node right;
    int value;
} node_t;

And that's how you do trees in C.

This won't work. The reason is that node has incomplete type at the point of declaration of left, i.e. the compiler doesn't know the size of node yet. Even worse (in the above code), this is a recursive composition, the compiler has to determine the size of node to determine the size of node.

The correct way is:

Code:

typedef struct node {
    struct node *left;
    struct node *right;
    int value;
} node_t;

The size of a pointer is independent of the referenced type, so the size of node is clear.

EAD · 04-15-2006, 03:42 AM

Tanx guys, nice info. I love the

Quote:

typedef struct node {
struct node *left;
struct node *right;
int value;
} node_t;

way, works nice.

taylor_venable · 04-15-2006, 12:57 PM

Quote:

Originally Posted by addy86

This won't work. The reason is that node has incomplete type at the point of declaration of left, i.e. the compiler doesn't know the size of node yet. Even worse (in the above code), this is a recursive composition, the compiler has to determine the size of node to determine the size of node.

Sorry about that; my mistake. In my hurry to point out the error from the typedef-ed node example, I ended up and made the same error myself. Just goes to show I should probably check my code more before I post it.

Thanks for pointing that out.

angustia · 04-15-2006, 10:36 PM

typedef struct node {
struct node *left;
struct node *right;
int value;
} node_t;

well, it's a good explanation, but it still sounds weird to explain to a newcomer...