Cast to function pointer

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 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.}

Later on I am casting, perhaps to call it like

 ((sigrout_t*) SGENT_1_calc) (someptr);