Returning function pointer type

Often I find the need to write functions which return function pointers. Whenever I do, the basic format I use is:

typedef int (*function_type)(int,int);

function_type getFunc()
{
   function_type test;
   test /* = ...*/;
   return test;
}

However this can get cumbersome when dealing with a large number of functions so I would like to not have to declare a typedef for each one (or for each class of functions)

I can remove the typedef and declare the local variable returned in the function as: int (*test)(int a, int b); making the function body look like this:

{
     int (*test)(int a, int b);
     test /* = ...*/;
     return test;
}

but then I do not know what to set for the return type of the function. I have tried:

int(*)(int,int) getFunc()
{
    int (*test)(int a, int b);
    test /* = ...*/;
    return test;
}

but that reports a syntax error. How do I declare the return type for such a function without declaring a typedef for the function pointer. Is it even possible? Also note that I am aware that it seems like it would be cleaner to declare typedefs, for each of the functions, however, I am very careful to structure my code to be as clean and easy to follow as possible. The reason I would like to eliminate the typedefs is that they are often only used to declare the retrieval functions and therefore seem redundant in the code.

int (*getFunc())(int, int) { … }

That provides the declaration you requested. Additionally, as ola1olsson notes, it would be good to insert void:

int (*getFunc(void))(int, int) { … }

This says that getFunc may not take any parameters, which can help avoid errors such as somebody inadvertently writing getFunc(x, y) instead of getFunc()(x, y).

You can probably do something like:

int foo (char i) {return i*2;}

int (*return_foo()) (char c)
{
   return foo;
}

but god, I hope I'll never have to debug you code....