I am learning Rust. And investigating examples. Why does the code outputs 1? My code:
struct S {
f: fn(),
}
impl S {
fn f(&self) {
print!("1");
}
}
fn main() {
let f2 = || print!("2");
let s = S{f: f2};
s.f(); // prints 1, but I expected 2
}
Rust distinguishes between methods defined on an impl block and fields associated with a struct. There is no relationship between the f in your S struct and the method whose full name is S::f which happens to take a &self argument.
s.f(); // Prints 1
This is syntax sugar for
S::f(&s); // Prints 1
It does not look at the fields of an object like Python does. When you have an expression followed by a method call syntax (i.e. a dot, a name, and an open parenthesis), that's always a method call. Method calls resolve statically and are treated as ordinary function calls by the runtime. s.f() could also refer to a method defined on a trait somewhere, if it were not defined in your inherent impl.
To call a function which happens to be stored in an instance variable, use extra parentheses.
(s.f)(); // Prints 2
This is not a method call. This is a field access, followed by calling the resulting field, which happens to be a function in this case.
