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Passing reference of reference to function argument (book v1)

In short, does passing reference A of another reference B to function parameter causing automatic dereference?

Here's the story...

When reading stack & heap - complex example, I am confused at the following explanation, please see img01 -> img02.

Since variable e is a reference to variable d, so the value of e is represented in -> 9 (i read it as: go to memory address of d), this is ok.

But why does the parameter f of baz get value of "data address referred by d", instead of the value of e? I thought the value of f should be -> 9 as well?

If f is what img02 says it is, then does it mean that "dereference" happened here automatically? Since e contains a reference to d, d contains a reference to some heap address, and the passing of e to f results in f has a reference to the heap address?

img01 img01

img02 img02

Solution

  • Yes, a dereference is done automatically. This is called Deref coercion.

    This is the mechanism that allows for implicitly converting one reference to another. Some notable uses:

    • &&T -> &T
    • &&&&&&&&T -> &T
    • &Box<T> -> &T
    • &Vec<T> -> &[T]

    See more explanation in the Rust book: Treating Smart Pointers Like Regular References with the Deref Trait or in the Rust reference: Type Coercions.