How to break down signal system call prototype in Linux?

Question

I am reading "The Linux Programming Interface", where I find signal system call prototype as stated:

#include <signal.h>
void ( *signal(int sig, void (*handler)(int)) ) (int);

What is signal here is it function-pointer or simple function?

I want to understand the reason behind putting last (int).

How come signal() is returning last desposition of SIGXXXX despite having void return type?

Dear @Biswapriyo, then how come it is returning last desposition of SIGXXXX despite having void return type? — MankPan, Sep 21 at 7:09
@Biswapriyo Michael Kerrisk (maintainer of the Linux man-pages project) wrote on his book that The return value of signal() is the previous disposition of the signal. Like the handler argument, this is a pointer to a function returning nothing and taking one integer argument. — MankPan, Sep 21 at 7:21
This is a question about C grammar, not related to Linux/Unix — 炸鱼薯条德里克, Sep 21 at 8:35
On stackoverflow: What's the meaning of this piece of code? void (*signal(int sig, void (*func)(int)))(int); — Mark Plotnick, Sep 21 at 12:12

Stephen Kitt · Accepted Answer · 2019-09-22 17:00:58Z

The way to read

void ( *signal(int sig, void (*handler)(int)) ) (int);

is that it declares a function, signal, which takes as arguments

an int
and a pointer to a function which itself takes an int and doesn’t return anything

and returns a pointer to a function which takes an int and doesn’t return anything.

The extra brackets correspond to the returned function pointer.

There’s a detailed explanation on Stack Overflow. Unfortunately cdecl.org doesn’t work on this declaration (because cdecl doesn’t either).

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