C type * (blank) declaration (binary tree, case in point)

Question

Here is a piece of code that's confusing me.

struct tNode {

  int key;            // search key for this item

  int data;           // data for this item

  struct tNode *left, *right; // children

};



typedef struct tNode tree_type;



/**************** tree_new() ****************/

/* Create a new tree */

tree_type *

tree_new(const int key, const int data)

{

  tree_type *node = malloc(sizeof(struct tNode));



  ...

  }

}

So I'm not quite getting the syntax for

tree_type *

right above tree_new function. Doesn't seem like a pointer... not sure what it's doing there.

--------------------- post-edit
Thanks to all who helped out!

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Answer 1

You have to read both lines together:

tree_type *
tree_new(const int key, const int data)

is just another legal way to write the function definition with its return type:

tree_type *tree_new(const int key, const int data)

since it doesn't end with a semicolon, the statement continues on the next line(s).

C type * (blank) declaration (binary tree, case in point)

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