Returning function pointer type

Question

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.

Answer 1

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).