Why is subsref and subscripted reference not equivalent

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Assume an struct array with the following content:
A = struct('b', {1, 2});
Following two code blocks do not give the same output:
ans =
1 2
S = struct('type', {'.'}, 'subs', {'b'});
[subsref(A, S)]
ans =
But I would say this two referencing methods are equal. What is the workaround for using subsref?

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Answers (2)

Kenneth Eaton
Kenneth Eaton on 21 Feb 2017
Not sure if this is a complete answer, but it appears that invoking subsref directly requires a stricter definition of the output arguments. While:
will automatically dump all possible output arguments in a comma separated list, calling subsref directly requires you to explicitly define how many output arguments you want. The only way I've figured out thus far to get them all is to collect them in a cell array of the appropriate size:
>> [out{1:numel(A)}]=subsref(A, S)
out =
1×2 cell array
[1] [2]
>> [out{:}]
ans =
1 2

Jan on 21 Feb 2017
Edited: Jan on 21 Feb 2017
Not an answer: I've played around with
S = struct('type', {'()', '.'}, 'subs', {{':'}, 'b'});
[subsref(A, S)]
is actually:
But this still replies 1 instead of the expected [1,2]. Then my interest in the question growed and I've voted for it. I expect, that there is a simple solution, by I cannot find it currently.


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