如何理解函数的函数参数中的星号（*）？

Question

Here is what I confused:

extern int sigsegv_leave_handler (
    void (*continuation) (void*, void*, void*), 
    void* cont_arg1, void* cont_arg2, void* cont_arg3);

I don't understand *continuation , what's the use for * in *continuation

update

The full code is in "/usr/include/sigsegv.h" in linux

update

I use sigsegv_leave_handler like following:

void cont(void *fault_addr, void *arg1, void *arg2) {
//    rb_raise(rb_eTypeError, "type err");
    rraise(SEGV, NULL);
}

int handle_segv(void *fault_addr, int serious) {
    sigsegv_leave_handler(cont, fault_addr, NULL, NULL);
}

Answer 1

This

void (*continuation) (void*, void*, void*)

is simply a declaration where continuation is a pointer to a function taking 3 void * pointers and not returning anything ( with void return type ).

This is how you declare pointers to functions,

RETURN_TYPE (*IDENTIFIER)(... PARAMETERS WITH THEIR TYPES AS USUAL ...);

Answer 2

A useful tool to use is C gibberish ↔ English

extern int sigsegv_leave_handler (void (*) (void*, void*, void*), void* , void* , void* );

declare sigsegv_leave_handler as extern function (pointer to function (pointer to void, pointer to void, pointer to void) returning void, pointer to void, pointer to void, pointer to void) returning int

Or in other words, sigsegv_leave_handler() takes 4 arguments and returns an int .

1) pointer to a function which takes 3 void * pointers and returns void
void (*f) (void*, void*, void*) 

2) pointer to void
void *

3) pointer to void
void *

4) pointer to void
void *

what's the use for * in *continuation?

To show that the argument is a pointer to a function.

void (*continuation) (void*, void*, void*),