What does these two arguments mean in typedef?

Question

I have a piece of code and I don't understand that one typedef:

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

Doesn't that actually mean you can use inst_cb_t as a void now? But what about the stuff in the second brackets?

Answer 1

The 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 use this only in function declaration and pointer to function deceleration.

inst_cb_t foo;  

You can't use it in function definition like

inst_cb_t foo      // WRONG
{
    // Function body
}  

Look at C standard:

C11: 6.9.1 Function definitions:

The identifier declared in a function definition (which is the name of the function) shall have a function type, as specified by the declarator portion of the function definition.¹⁶²⁾

and footnote 162 is

The intent is that the type category in a function definition cannot be inherited from a 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