How to create an empty dictionary with a struct as value

I want to create a dictionary with a defined struct as the value. What i currently have is:

dict = dictionary(string.empty, struct(test={}));

The problem is that the empty dictionary throws an error when i try to access the struct it throws an error that the field name does not exist:

dict.values.test

Unrecognized field name "test".

I want that the dictionary behave like when you try to access an empty struct with a field. Is there any way so the dictionary remembers the field values?

[Jan]

What do you expect struct(test={}) to do? Why do you think, that this creates a struct called "test"?

[Guido Hiller]

I expect it to create an empty struct which has the field test.

str = struct(test={});

I can then access the field of the empty struct without it throwing errors and just returning nothing.

str.test

As far as I understand it, when I use the struct in the dictionary it erases the field (?), so dict.values.test returns an error.

[Rik]

@Jan, As you can see below, it actually does something, as this is using the Name=Value syntax. So it is equivalent to the second call:

str = struct(test={})

str = 

  0×0 empty struct array with fields:

    test

str = struct('test',{})

str = 

  0×0 empty struct array with fields:

    test

[Jan]

@Rik: Thanks. I ran the same code struct(test={}) yesterday here in the forum and got an error message. Strange.

@Guido Hiller: Thanks for your explanation. What is the meaning of assining an empty string in a dictionary? What about considering as a bug, that the dictionary command acceptes an empty value for an empty key? Would stopping with an error message be better?

Or in other words: Is an empty dictionary meaningful? I know, there are empty structs with fields, so maybe the answer is "yes".

[Guido Hiller]

@Jan The dictionary is filled based on user input and there is a legit case that the dictionary is just empty. Basically it is just me being to lazy to write an if else clause everytime around the getter.

Another methode i am currently testing is creating a custom class which contains just the properties from the struct, which seems to work better with the dictionary.

[Paul]

@Guido Hiller

I still don't understand what the use case is for having an empty, configured dictionary where the value of the empty dictionary is a struct with a field. What would you do with that empty dictionary, which I guess is the same question asked by @Jan.

If we configure a dictionary with an empty string key and empty struct value

d = dictionary(string.empty,struct.empty)

d = 
  dictionary (string ⟼ struct) with no entries.

we can subsequently add entries to the dictionary

d("a") = struct(test=1);
d("b") = struct(test=2);

d.values returns a struct array

d.values

ans = 2×1 struct array with fields:
    test

which, in this case can be accessed as a struct and the values in the test field turned into an array via a comma separated list in the usual way

[d.values.test]

ans = 1×2

1 2

We can add an entry with a different type of value in the test field

d("c") = struct(test="foo")

d = 
  dictionary (string ⟼ struct) with 3 entries:

    "a" ⟼ 1×1 struct
    "b" ⟼ 1×1 struct
    "c" ⟼ 1×1 struct

d.values

ans = 3×1 struct array with fields:
    test

d.values.test

ans = 1

ans = 2

ans = "foo"

We can even add a struct with a different field

d("d") = struct(other=100)

d = 
  dictionary (string ⟼ struct) with 4 entries:

    "a" ⟼ 1×1 struct
    "b" ⟼ 1×1 struct
    "c" ⟼ 1×1 struct
    "d" ⟼ 1×1 struct

But now we can't access values as above because the struct values are different and can't themeselves be concateneated into a struct array

try
    d.values
catch ME
    disp('error')
end

error

But we can still get the values in a cell array

values(d,"cell")

ans = 4×1 cell array
    {1×1 struct}
    {1×1 struct}
    {1×1 struct}
    {1×1 struct}

[Guido Hiller]

@Paul Thank you for your answer. I think I had a misunderstanding about the behaviour of structs. If you can put structs with different fields into the same dictionary, then my inital question does not make any sense.