指向函数的指针的语法，该函数返回C中的函数指针

Question

How to declare a pointer to a function returning another function pointer? Please share with me the syntax and an example code snippet.

Also, in which scenario would a function pointer returning a function pointer be used?

Answer 1

This is trivial with typedefs:

typedef int(*FP0)(void);
typedef FP0(*FP1)(void);

FP1 is the type of a pointer to a function that returns a function pointer of type FP0 .

As for when this is useful, well, it is useful if you have a function that returns a function pointer and you need to obtain or store a pointer to this function.

Answer 2

If you avoid using typedef , it is hard. For example, consider signal() from the C standard:

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

void handler(int signum)
{
    ...
}

if (signal(SIGINT, SIG_IGN) != SIG_IGN)
    signal(SIGINT, handler);

Using typedefs, it is easier:

typedef void Handler(int);
extern Handler *signal(int, Handler *);

void handler(int signum)
{
    ...
}

if (signal(SIGINT, SIG_IGN) != SIG_IGN)
    signal(SIGINT, handler);

Note that for the signal() function, you would normally simply use <signal.h> and let the system worry about declaring it.

Answer 3

If you don't want a second typedef,

typedef float(*(*fptr_t)(int))(double)

this means " declare fptr_t as pointer to function (int) returning pointer to function (double) returning float " (fptr_t : int → (double → float))