C function pointer declaration omitting parameters

In the code, I omitted parameters for int (*bar) and assigned &foo, which as 3 arguments, to bar. (*bar) received the numbers and gave me a return value. I thought this is ok but I've heard that this is actually UB. How does (*bar) receive the numbers? Thx

#include <stdio.h>

int foo(int a, int b, int c){
    return a+b+c;
}

int main(void) {
    int (*bar)() = &foo;
    printf("%d", bar(1, 2, 3));
    return 0;
}

edit When I pass more than three arguments to bar() (say 5 arguments), the program works. Where do the two extra arguments go?