if this structure is to be used by some other file func.c how to do it?

When a type is used in a file (i.e. func.c file), it must be visible. The very worst way to do it is copy paste it in each source file needed it.

The right way is putting it in an header file, and include this header file whenever needed.

shall we open a new header file and declare the structure there and include that header in the func.c?

This is the solution I like more, because it makes the code highly modular. I would code your struct as:

#ifndef SOME_HEADER_GUARD_WITH_UNIQUE_NAME
#define SOME_HEADER_GUARD_WITH_UNIQUE_NAME

struct a
{ 
    int i;
    struct b
    {
        int j;
    }
};

#endif

I would put functions using this structure in the same header (the function that are "semantically" part of its "interface").

And usually, I could name the file after the structure name, and use that name again to choose the header guards defines.

If you need to declare a function using a pointer to the struct, you won't need the full struct definition. A simple forward declaration like:

struct a ;

Will be enough, and it decreases coupling.

or can we define the total structure in header file and include that in both source.c and func.c?

This is another way, easier somewhat, but less modular: Some code needing only your structure to work would still have to include all types.

In C++, this could lead to interesting complication, but this is out of topic (no C++ tag), so I won't elaborate.

then how to declare that structure as extern in both the files. ?

I fail to see the point, perhaps, but Greg Hewgill has a very good answer in his post How to declare a structure in a header that is to be used by multiple files in c?.

shall we typedef it then how?

• If you are using C++, don't.
• If you are using C, you should.

The reason being that C struct managing can be a pain: You have to declare the struct keyword everywhere it is used:

struct MyStruct ; /* Forward declaration */

struct MyStruct
{
   /* etc. */
} ;

void doSomething(struct MyStruct * p) /* parameter */
{
   struct MyStruct a ; /* variable */
   /* etc */
}

While a typedef will enable you to write it without the struct keyword.

struct MyStructTag ; /* Forward declaration */

typedef struct MyStructTag
{
   /* etc. */
} MyStruct ;

void doSomething(MyStruct * p) /* parameter */
{
   MyStruct a ; /* variable */
   /* etc */
}

It is important you still keep a name for the struct. Writing:

typedef struct
{
   /* etc. */
} MyStruct ;

will just create an anonymous struct with a typedef-ed name, and you won't be able to forward-declare it. So keep to the following format:

typedef struct MyStructTag
{
   /* etc. */
} MyStruct ;

Thus, you'll be able to use MyStruct everywhere you want to avoid adding the struct keyword, and still use MyStructTag when a typedef won't work (i.e. forward declaration)

Edit:

Corrected wrong assumption about C99 struct declaration, as rightfully remarked by Jonathan Leffler.

Edit 2018-06-01:

Craig Barnes reminds us in his comment that you don't need to keep separate names for the struct "tag" name and its "typedef" name, like I did above for the sake of clarity.

Indeed, the code above could well be written as:

typedef struct MyStruct
{
   /* etc. */
} MyStruct ;

IIRC, this is actually what C++ does with its simpler struct declaration, behind the scenes, to keep it compatible with C:

// C++ explicit declaration by the user
struct MyStruct
{
   /* etc. */
} ;
// C++ standard then implicitly adds the following line
typedef MyStruct MyStruct;

Back to C, I've seen both usages (separate names and same names), and none has drawbacks I know of, so using the same name makes reading simpler if you don't use C separate "namespaces" for structs and other symbols.

Caveat: It is not necessary to put the implementation in the header file, see the alternative solution at the end of this answer.

Anyway, the reason your code is failing is that, when instantiating a template, the compiler creates a new class with the given template argument. For example:

template<typename T>
struct Foo
{
    T bar;
    void doSomething(T param) {/* do stuff using T */}
};

// somewhere in a .cpp
Foo<int> f; 

When reading this line, the compiler will create a new class (let's call it FooInt), which is equivalent to the following:

struct FooInt
{
    int bar;
    void doSomething(int param) {/* do stuff using int */}
}

Consequently, the compiler needs to have access to the implementation of the methods, to instantiate them with the template argument (in this case int). If these implementations were not in the header, they wouldn't be accessible, and therefore the compiler wouldn't be able to instantiate the template.

A common solution to this is to write the template declaration in a header file, then implement the class in an implementation file (for example .tpp), and include this implementation file at the end of the header.

Foo.h

template <typename T>
struct Foo
{
    void doSomething(T param);
};

#include "Foo.tpp"

Foo.tpp

template <typename T>
void Foo<T>::doSomething(T param)
{
    //implementation
}

This way, implementation is still separated from declaration, but is accessible to the compiler.

Alternative solution

Another solution is to keep the implementation separated, and explicitly instantiate all the template instances you'll need:

Foo.h

// no implementation
template <typename T> struct Foo { ... };

Foo.cpp

// implementation of Foo's methods

// explicit instantiations
template class Foo<int>;
template class Foo<float>;
// You will only be able to use Foo with int or float

If my explanation isn't clear enough, you can have a look at the C++ Super-FAQ on this subject.