Clarification on function pointers in C [duplicate]

Question

The following code comes from example abo3.c from Insecure Programming — see also Why cast extern puts to a function pointer (void(*)(char*))&puts?:

int main(int argv,char **argc) {
    extern system,puts; 
    void (*fn)(char*)=(void(*)(char*))&system; // <==
    char buf[256];
    fn=(void(*)(char*))&puts;
    strcpy(buf,argc[1]);
    fn(argc[2]);
    exit(1);
}

Specifically this line:

void (*fn)(char*)=(void(*)(char*))&system;

I think that void (*fn)(char*) sounds like a lambda, but I know that it's not. Then, maybe this is only a play with parentheses, where void *fn(char*) is a declaration of a function and this function is referencing system? But why does the (char*) parameter have no name? Is this permitted?

Answer 1

It declares the variable fn as a function pointer (to a function that has one argument of type char * and does not return anything (void).

This variable is initialised with the address of system - see http://linux.die.net/man/3/system. As noted from this page this will require the cast as given