在c中写入一个函数指针 [英] Writing a function pointer in c
问题描述
int(* fn_pointer(this_args(this_args))这个函数指针被写为: ))(this_args)
我通常遇到这样的函数指针:
return_type(* fn_pointer)(arguments);
讨论类似的事情:
/ /这是一个名为functionFactory的函数,它接收参数n
//并返回一个指向另一个函数的指针,该函数接收两个ints
//并返回另一个int
int(* functionFactory(int n ))(int,int){
printf(Got parameter%d,n);
int(* functionPtr)(int,int)=& addInt;
返回functionPtr;
}
有人可以告诉我有什么不同,以及它是如何工作的? p>
int(* fn_pointer(this_args))(this_args);
声明 fn_pointer
是一个函数, code> this_args 并返回一个指向一个函数的指针,该函数以 this_args
作为参数,并返回一个 int
类型。它相当于
typedef int(* func_ptr)(this_args);
func_ptr fn_pointer(this_args);
让我们多了解一下:
int f1(arg1,arg2); // f1是一个函数,它接受两个类型为
// arg1和arg2的参数并返回一个int。
int * f2(arg1,arg2); // f2是一个函数,它接受两个类型为
// arg1和arg2的参数并返回一个指向int的指针。
int(* fp)(arg1,arg2); // fp是一个指向函数的指针,该函数接受两个类型为
// arg1和arg2的参数,并返回一个指向int的指针。
int f3(arg3,int(* fp)(arg1,arg2)); // f3是一个函数,它接受
的两个参数//类型arg3和一个函数指针
//接受两个类型为arg1和arg2的参数和
// //返回一个int 。
int(* f4(arg3))(arg1,arg2); // f4是一个函数,它接受
// arg3类型的参数并返回一个指向一个函数的指针,该函数接受两个
//类型为arg1和arg2的参数并返回一个int
如何阅读 int(* f4(arg3))(arg1,arg2);
f4 - f4
f3() - 是一个函数
f3(arg3) - 接受一个arg3参数
* f3(arg3) - 返回一个指针
(* f3(arg3))() - 函数
(* f3(arg3))(arg1,arg2) - 取arg1和arg2参数
int(* f3(arg3))( arg1,arg2) - 并返回一个int
所以,最后一个家庭作品:)。尝试找出声明
void(* signal(int sig,void(* func)(int)))(int );
并使用 typedef
重新定义它。 / p>
I was recently reading a code, and found that a function pointer is written as :
int (*fn_pointer ( this_args ))( this_args )
I usually encounter a function pointer like this :
return_type (*fn_pointer ) (arguments);
Similar thing is discussed here:
// this is a function called functionFactory which receives parameter n
// and returns a pointer to another function which receives two ints
// and it returns another int
int (*functionFactory(int n))(int, int) {
printf("Got parameter %d", n);
int (*functionPtr)(int,int) = &addInt;
return functionPtr;
}
Can somebody tell me what is the difference and how does this work ?
int (*fn_pointer ( this_args ))( this_args );
declares fn_pointer
as a function that takes this_args
and returns a pointer to a function that takes this_args
as argument and returns an int
type. It is equivalent to
typedef int (*func_ptr)(this_args);
func_ptr fn_pointer(this_args);
Let's understand it a bit more:
int f1(arg1, arg2); // f1 is a function that takes two arguments of type
// arg1 and arg2 and returns an int.
int *f2(arg1, arg2); // f2 is a function that takes two arguments of type
// arg1 and arg2 and returns a pointer to int.
int (*fp)(arg1, arg2); // fp is a pointer to a function that takes two arguments of type
// arg1 and arg2 and returns a pointer to int.
int f3(arg3, int (*fp)(arg1, arg2)); // f3 is a function that takes two arguments of
// type arg3 and a pointer to a function that
// takes two arguments of type arg1 and arg2 and
// returns an int.
int (*f4(arg3))(arg1, arg2); // f4 is a function that takes an arguments of type
// arg3 and returns a pointer to a function that takes two
// arguments of type arg1 and arg2 and returns an int
How to read int (*f4(arg3))(arg1, arg2);
f4 -- f4
f3( ) -- is a function
f3(arg3) -- taking an arg3 argument
*f3(arg3) -- returning a pointer
(*f3(arg3))( ) -- to a function
(*f3(arg3))(arg1, arg2) -- taking arg1 and arg2 parameter
int (*f3(arg3))(arg1, arg2) -- and returning an int
So, finally a home work :). Try to figure out the declaration
void (*signal(int sig, void (*func)(int)))(int);
and use typedef
to redefine it.
这篇关于在c中写入一个函数指针的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!