cell array expansion

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Daniel Shub
Daniel Shub on 21 Jul 2011
I have just been bitten by some careless coding, but I am surprised that MATLAB lets me do it and that mlint didn't provide a warning. Why does MATLAB let you do this:
x = {1,2};
y = x{:};
what I wanted to do (and I am sure there are a number of other ways of doing it) was
y = [x{:}];
I can see the advantage of cell array expansion for referencing and parameter passing. For example,
z = magic(5);
z(x{:})
xy = {1:10, 0:9};
plot(xy{:});
Is there any reason for
y = x{:};
to be valid. I feel like it should return an error about the RHS returning more arguments than the LHS.

Accepted Answer

Jan
Jan on 21 Jul 2011
This is the standard behaviour of Matlab:
x = {1, 2};
y = x{:}; % y is x{1}
You see the same method for:
a = rand(1, 10);
b = max(a);
Why is this equivalent? Because MAX replies 2 outputs as "x{:}" and if just the 1st is caught, the 2nd is ignored. Therefore these methods are equivalent also:
x = {1, 2};
[y1, y2] = x{:}
a = rand(1, 10);
[b1, b2] = max(a)
So I would not expect an MLint-warning for a standard behaviour.
  1 Comment
Daniel Shub
Daniel Shub on 6 Aug 2011
I think you explanation is the best, even if I don't fully agree. I see a difference if you consider max(a) and x{:} without semicolons (max returns one answer and x returns two).

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More Answers (2)

Sean de Wolski
Sean de Wolski on 21 Jul 2011
One reason: For a function that takes varargin you can feed it all contents of a cell as a separate input.
x = {1,2,3,4}
cat(3,x{:})
Yes, it frustrates me some times too!
  3 Comments
Daniel Shub
Daniel Shub on 21 Jul 2011
I don't understand. Yes, x = {1, [1, 2]}; y = [x{:}]; will fail and y = x{:} won't fail, but why would you want y = x{:}?

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Fangjun Jiang
Fangjun Jiang on 21 Jul 2011
Maybe one way to explain it is to treat it as the variable output argument. Like,
MaxValue=max(1:10);
[MaxValue,Index]=max(1:10);
You can do:
x={1,2};
y=x{:}
[a b]=x{:}

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