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Normally, whenever a function designator (a function's name) is used in an expression, it decays to a function pointer to that function. 

`typeof` is one of those exceptions to the "decay" rules of C (C23 6.3.2.1 §4), so if you give it a function type as operand, the result is a function type and not a function pointer type.

So it is now possible to write obscure declarations like:

```c
void a (void);
typeof(a) b;
```

And that's equivalent to:

```c
void a (void);
void b (void);
```

So therefore given some function `int foo (FILE*, FILE*)`, then`typeof(foo)*` will yield a function pointer of the type `int (*) (FILE*, FILE*)`. Meaning that `operation` and `op` in your example would indeed be equivalent.

I would however not recommend this style since it is obscure and also not backwards-compatible. The clearest style IMO, compatible all the way back to C90 is this:

    typedef int foo_t (FILE*, FILE*); // typedef a function not a pointer
    ...
    foo_t* fp1;       // function pointer
    foo_t* const fp2; // function pointer residing in ROM

This makes function pointer syntax 100% in sync with object pointer syntax.