c++ - What does void(*)() mean in code -

i saw code today in fb profile, , not able understand , how working:-

(*(void(*)()) shellcode)() 

can please explain me, above code mean ?

full code snippet below :-

#include <stdio.h> #include <string.h>  char *shellcode = "\x31\xc0\x50\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69"           "\x6e\x89\xe3\x50\x53\x89\xe1\xb0\x0b\xcd\x80";  int main(void) { fprintf(stdout,"length: %d\n",strlen(shellcode)); (*(void(*)()) shellcode)(); return 0; } 

it cast function pointer (with no returned result , no arguments). prefer using typedef define signature of such functions:

 typedef void plainsig_t(void); 

then code

 (*(plainsig_t*)shellcode) (); 

for function pointers, don't need dereference them, shorter code:

 ((plainsig_t*) shellcode) (); 

which calls function machine code located inside shellcode memory zone.

btw, not strictly portable c. in principle, there no guarantee can cast data pointer function pointer. (on weird processors -e.g. embedded microcontrollers, dsp, 1970s era computers-, code , data sit in different address spaces, or have different pointer sizes, etc....). common processors , abi (x86-64/linux, arm/android, ....) have same address space code , data , accept casting function pointers data pointers , vice versa.