What do these two arguments mean in typedef?

Question

I have a piece of code, and I don’t understand what one typedef :

 typedef void (inst_cb_t) (const char*, size_t); 

Does this mean that now you can use inst_cb_t as void ? But what about the material in the second brackets?

Answer 1

Declaration

 typedef void (inst_cb_t) (const char*, size_t); 

defines inst_cb_t as a function that takes two arguments of type const char* and size_t and returns void .

One interesting part of this declaration is that you can only use this in a function declaration and a pointer to slowing down a function.

 inst_cb_t foo; 

You cannot use it in a function definition like

 inst_cb_t foo // WRONG { // Function body } 

Take a look at standard C:

C11: 6.9.1 Function definitions:

The identifier declared in the function definition (which is the name of the function) must be of the function type, as indicated in the declaration part of the function definition. ¹⁶²⁾

and footnote 162 -

The intention is that the type category in a function definition cannot be inherited from typedef :

 typedef int F(void); // type F is ''function with no parameters // returning int F f, g; // f and g both have type compatible with F F f { /* ... */ } // WRONG: syntax/constraint error F g() { /* ... */ } // WRONG: declares that g returns a function int f(void) { /* ... */ } // RIGHT: f has type compatible with F int g() { /* ... */ } // RIGHT: g has type compatible with F F *e(void) { /* ... */ } // e returns a pointer to a function F *((e))(void) { /* ... */ } // same: parentheses irrelevant int (*fp)(void); // fp points to a function that has type F F *Fp; // Fp points to a function that has type F