Have come across the line of code shown below I think it may be a cast to a function pointer that returns void and takes a void pointer, is that correct?
(void (*)(void *))SGENT_1_calc
Yes it is correct. I find that not very readable, so I suggest declaring the signature of the function to be pointed:
typedef void sigrout_t(void*);
I also have the coding convention that types ending with rout_t are such types for functions signatures. You might name it otherwise, since _t is a suffix reserved by Posix
latter on I am casting, perhaps to call it like
((sigrout_t*) SGENT_1_calc) (someptr);
Yes it is, the function should be looking like this
void func(void*);
But the statement is missing a target, since a cast to nothing is useless. So it should be like
func = (void (*)(void *))SGENT_1_calc;
Yes it is a cast as you have stated.
yes its a function pointer which you can assign to a function with proto void funcname(void*) Here the SGENT_1_calc can be directly assigned to funcname
来源：https://stackoverflow.com/questions/15807333/cast-to-function-pointer