Pointer in typedef struct confusion

Question

I am trying to define a typedef struct as follows:

typedef struct node{ int keys[2*MIN_DEGREE-1]; struct node* child[2*MIN_DEGREE]; int numkeys; int isleaf; } BNODE,*BNODEPTR;

Instead of using struct node* child[2*MIN_DEGREE] why can't I declare the struct as follows:

typedef struct node{ int keys[2*MIN_DEGREE-1]; BNODEPTR child[2*MIN_DEGREE]; int numkeys; int isleaf; } BNODE,*BNODEPTR;

I am little confused as to how the compiler resolves structs that has nested pointers to the same type. It will be great somebody helps me clear this up.

Thanks

Jonathan Leffler · Accepted Answer · 2014-03-23 01:24:45Z

You can't use BNODEPTR in the body of the structure like that because it either doesn't exist as a type at all until after the definition after the close brace of the structure body, or (worse) it refers to a different type altogether^*.

You could use:

typedef struct node BNODE, *BNODEPTR; struct node { int keys[2*MIN_DEGREE-1]; BNODEPTR child[2*MIN_DEGREE]; int numkeys; int isleaf; };

And there's another whole argument that says BNODEPTR is evil and you should only use BNODE and BNODE *, but that's a style issue, not a technical one.

Were it my code, it would probably be more like:

typedef struct Node Node; struct Node { int keys[2*MIN_DEGREE-1]; Node *child[2*MIN_DEGREE]; int numkeys; int isleaf; };

In C++, the rules are slightly different and you would not need the typedef line (so Node would be known as a type from the open brace).

^{* This can only happen if the original BNODEPTR is defined at an outer scope and this one appears inside a function, but when it happens, it is really confusing!}

Jonathan Leffler · Accepted Answer · 2014-03-23 01:40:29Z

Instead of using struct node* child[2*MIN_DEGREE] why can't I declare the struct as follows: BNODEPTR child[2*MIN_DEGREE];?

At that point, compiler (yet) does not know what the symbol BNODEPTR is.

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