Unexpected behavior of anonymous function

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The anonymous function k below behaves correcltly except for the last two cases k(1,1,:) and k(1,2,:), where it interprets the semicolon as a charcter (':'=58, 58^2=3364), while it should return the handle in the first case and error in the second. Any explanations?
>> k=@(varargin) cellfun(@(x) x^2,varargin)
k =
>> k(:)
ans =
>> k(1)
ans =
>> k(1,:)
ans =
>> k(1,2)
ans =
1 4
>> k(1,1,:)
ans =
1 1 3364
>> k(1,2,:)
ans =
1 4 3364
Cedric Wannaz
Cedric Wannaz on 24 Jan 2013
Edited: Cedric Wannaz on 24 Jan 2013
@Sean: yes, it is as if when S.subs is larger than 2, subs are not treated the same way.. and it is not the position of ':' in the subs that matters:
>> k(:,1,2)
ans = 3364 1 4

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Accepted Answer

Steve Eddins
Steve Eddins on 24 Jan 2013
First, I would like to point out that k(1,1,:) is a valid expression for a subscripting operation on a variable called k, but it is not a valid expression for a call to a function called k.
In other words, you can't pass a "naked" colon as a function argument!
>> sin(:)
Undefined variable sin.
So k(:), k(1,:), k(1,1,:), and k(1,2,:) are all invalid ways to call a function (or a function handle).
The fact that MATLAB isn't just giving a quick error on these expressions is an artifact of the way function handle evaluation syntax using parentheses has been implemented. It's been implemented by overloading the parentheses syntax using subsref. By the time the subsref overload sees things, the "naked" colon has been converted to a character. Like everything else inside the parentheses, this character then gets treated as a function argument.
  1 Comment
Khaled Hamed
Khaled Hamed on 24 Jan 2013
I do agree. It just seemed logical that k(:), k(1,:), k(1,1,:), and k(1,2,:) would be interpreted as indexing (the first three would work for a scalar, while the fourth would error).

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More Answers (1)

Sean de Wolski
Sean de Wolski on 24 Jan 2013
Please contact us and reference this thread.
That certainly looks like obscure behavior.


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