What exactly happens when you typedef a function pointer b/c my test code works but I don't think it should?

I am learning C and C++. I wrote a program in C++ and I am writing the same program in C. I want to use structs in C to mimic the classes I wrote in C++. To this end, I wrote test code that (1) creates a typedef to a function pointer, (2) a struct with the typedef as a member, and (3) assigns the member to a compatible function.

#include <stdio.h>

// comment 1 -- see next line
typedef int (*type_def_func_ptr) (int, int);

int result(int x, int y){
    int temp_result = x + y;
    return temp_result;
}

struct TEST_STRUCT {
    int day;
    int month;
    int year;
    type_def_func_ptr tdfp;
};

int main(){
    struct TEST_STRUCT MyStruct;
    MyStruct.day = 28;
    MyStruct.month = 9;
    MyStruct.year = 2024;
        // comment 2 -- see next line
    MyStruct.tdfp = result; 
    int result1 = MyStruct.tdfp(28, 9);
    printf("Result1 is %d\n", result1);
}

The code works. I don't understand, however, why the code works as written without (1) result() being *result() (comment 1) or (2) without MyStruct.tdfp = result being = &result (comment 2). The code still works with (2) being &result or *result, which leads me to believe that the compiler is interpreting what I wrote instead of blindly compiling what I wrote.

1. int result1 = MyStruct.tdfp(28, 9); and int result1 = (*MyStruct.tdfp)(28, 9); do the same thing.

2. MyStruct.tdfp = result; and MyStruct.tdfp = &result; also do the same thing.

These convenience "shortcuts" do the same thing as in C++ (in which you only need & to take the address of functions if they are member functions).