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Are these two function pointer declarations equivalent?
Say I have two functions:
FILE* get_input(const char fname[static 1]);
FILE* get_output(const char fname[static 1]);
And I wish to declare a function pointer and assign it the result of some processing which shall be called later.
Are these two declarations:
int foo(FILE *, FILE *);
int (*operation)(FILE *, FILE *);
typeof (foo) *op;
equivalent? Or is there any difference between the types of operation and op? If so, how do I properly declare a function pointer with typeof?
1 answer
The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| Melkor-1 | (no comment) | Jun 25, 2024 at 14:23 |
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:
void a (void);
typeof(a) b;
And that's equivalent to:
void a (void);
void b (void);
So therefore given some function int foo (FILE*, FILE*), thentypeof(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.
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