cell array expansion

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.

Answer 1

[Jan]

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.

[Daniel Shub]

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).

Answer 2

[Sean de Wolski]

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!

[Daniel Shub]

Yes, I agree and this is basically what I pointed out in my plot(xy{:}) example. What I want to know is if y = x{:} is useful. If the only reason it is valid is so that f(x{:}) can work then I think mlint should provide a warning (and even better MATLAB should throw an error).

[Sean de Wolski]

I disagree; that functionality is quite useful. What if the cell contains different sized elements? Concatenating them as you have done will fail.

[Daniel Shub]

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{:}?

Answer 3

[Fangjun Jiang]

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{:}