This code is just a situation that I found in my actual code which is very big so I\'m giving this.here in this code the structure \"struct node\" is not defined it
This has nothing to do with the extern
keyword on the prototype (although it's not necessary).
You need a forward declaration for struct node
:
/* test2.h */
struct node; // <== this
extern void f(struct node * k);
Without that forward declaration, the struct node
inside the function prototype is local to that particular function prototype.
So when the compiler sees the definition of f()
it also sees another different local declaration for a struct node
and thinks that you have a conflicting declaration for f()
.
See C99 6.9.1 "Scopes of identifiers":
For each different entity that an identifier designates, the identifier is visible (i.e., can be used) only within a region of program text called its scope. Different entities designated by the same identifier either have different scopes, or are in different name spaces. There are four kinds of scopes: function, file, block, and function prototype. (A function prototype is a declaration of a function that declares the types of its parameters.)
...
If the declarator or type specifier that declares the identifier appears within the list of parameter declarations in a function prototype (not part of a function definition), the identifier has function prototype scope, which terminates at the end of the function declarator.
GCC 4.6.1 gives a nice warning about this for me:
C:\temp\test.c:3:22: warning: 'struct node' declared inside parameter list [enabled by default]
C:\temp\test.c:3:22: warning: its scope is only this definition or declaration, which is probably not what you want [enabled by default]
C++ makes function prototype scope apply only to the parameter names (C++03 3.3.3), so you won't see this problem if you compile the code as C++.