How could I access the resource in a resource type？

Move
#1

I know that by borrowing and dereference I could access the value of the field of a resource. But I can’t use this method to access a resource in a resource. In the following code, fun1 can access a resource’s variable (u64) , but fun2 cannot access that resource’s variable(resource) .

resource T1 {
	a: u64
}
resource T2 {
	t1: T1,
	b: u64
}
//ok
public fun1(): u64 acquires T2
{
    let add: address;
    let ref: &mut Self.T2;     
    let x: u64;   
    add = get_txn_sender();
    ref = borrow_global_mut<T2>(copy(add));
    x = *&move(ref).b;
    return move(b);
}
//error
public fun2(): Self.T1 acquires T2
{
    let add: address;
    let ref: &mut Self.T2;     
    let t: T1;   
    add = get_txn_sender();
    ref = borrow_global_mut<T2>(copy(add));
    t = *&move(ref).t1;
    return move(t);
}

So how could I access the field (resource) in a resource type？

#2

fun2 is using the correct scheme for accessing a child resource: acquire a reference to the parent resource, then acquire a reference to the field that holds the child resource.

The problem with fun2 is that it’s copying a resource via t = *&move(ref).t1. This statement is actually doing two things:

  • acquire a reference to the t1 field of t (ok)
  • dereference it (not ok, because it copies the resource)

The things you can do with the reference to a resource (e.g., t1) are:

  • Call a procedure that takes a &Self.T1
  • Acquire a reference to a field of the Self.T1 resource (but only because T1 is defined in the current module; this wouldn’t work if T1 were a resource defined elsewhere)
  • Return it (though not in this case, because it’s a reference to a global. This would be allowed if ref were a formal of fun2).

Hope this is useful! If you share some context on what you are trying to do in fun2, we can try to help.

1 Like
#3

Thank you very much for your answer. It helped me a lot.
I’ve summarized several resource usage scenarios as follows:

  1. access a field (u64) in a resource
module T
{
    resource T1
    {
        a: u64
    }
    resource T2
    {
        b: u64,
        t1: Self.T1
    }
    public fun1(): u64 acquires T1
    {
        let add: address;
        let ref: &mut Self.T1;
        let x: u64;
        add = get_txn_sender();
        ref = borrow_global_mut<T1>(move(add));
        x = *&move(ref).a;
        return move(x);
    }
}
  1. modify a field(u64) in a resource
module T
{
    resource T1
    {
        a: u64
    }
    resource T2
    {
        b: u64,
        t1: Self.T1
    }
    public fun2() acquires T1
    {
        let add: address;
        let ref: &mut Self.T1;
        let x: u64;
        add = get_txn_sender();
        ref = borrow_global_mut<T1>(move(add));
        x = *&copy(ref).a;
        *(&mut move(ref).a) = move(x) + 1;
        return;
    }
}
  1. access a field (u64) in a child resource of the resource
module T
{
    resource T1
    {
        a: u64
    }
    resource T2
    {
        b: u64,
        t1: Self.T1
    }
    public fun3(): u64 acquires T2
    {
        let add: address;
        let ref: &mut Self.T2;     
        let t: &mut Self.T1;   
        let x: u64;  
        add = get_txn_sender();
        ref = borrow_global_mut<T2>(move(add));
        t = &mut move(ref).t1;
        x = *&copy(t).a;
        return move(x);
    }
}
  1. modify a field (u64) in a child resource of a parent resource
module T
{
    resource T1
    {
        a: u64
    }
    resource T2
    {
        b: u64,
        t1: Self.T1
    }
    public fun4() acquires T2
    {
        let add: address;
        let ref: &mut Self.T2;     
        let t: &mut Self.T1;   
        let x: u64;  
        add = get_txn_sender();
        ref = borrow_global_mut<T2>(move(add));
        t = &mut move(ref).t1;
        x = *&copy(t).a;
        *(&mut move(t).a) = move(x) + 1;
        return;
    }
}

Is my method correct?
And when the child resource isn’t in the same module with the parent resource, like follow, what could I do to access the child resource and its member variables?

module Tt
{
    resource T3
    {
        c: u64
    }
}
module T
{
    import Transaction.Tt;
    resource T1
    {
        a: u64
    }
    resource T2
    {
        b: u64,
        t1: Self.T1,
        t3: Tt.T3
    }
    public fun5()
    {
        //I want to access or modify the value of c in resource T3 
    }
}
1 Like
#4

Is my method correct?

Yes, the examples above all look correct to me.

And when the child resource isn’t in the same module with the parent resource, like follow, what could I do to access the child resource and its member variables?

A field f of a resource S can only be borrowed inside the module that declares S (i.e., “all fields are private”). This means that if you want to access f outside of S, you can only do it by calling a procedure of S.

#5

Thank you very much. With your help, I successfully realized the function of accessing the child resource in different module.
And I still have a question: If I publish a resource T to address add , does that mean I publish the child resource of T to add at the same time?

#6

And I still have a question: If I publish a resource T to address add , does that mean I publish the child resource of T to add at the same time?

If I understand your question correctly, the answer is yes. Each address has at most one resource of a given type published at the top level. However, top level resources can themselves store child resources.

For example, an account could store both a T1 and a T2 resource at the top level. The top-level T1 is distinct from the child T1 stored in T2.

1 Like
#7

So, if I publish a resource T1 which has a child resource T2 to an account, could I get T2 from the account by borrow_global or move_from?

#8

No. The T2 child can only be accessed via (1) a borrow_global<T1> followed by a borrow of the field of type T2, or (2) a move_from<T1> followed by an unpack to push the contents of T1 (including the field of type T2) onto the stack).

#9

I got it. Thank you very much for your answer!