Using AND Operator in “if” statements

Hi,

When I type the following code:

if size([1 2 3])==size([4 5 6]) & size([4 5 6])==size([7 8 9])
    'yes'
else
    'no'
end

MATLAB Code Analyzer issues this warning message: "When both arguments are numeric scalars, consider replacing & with && for performance."

So, I use && instead of &:

if size([1 2 3])==size([4 5 6]) && size([4 5 6])==size([7 8 9])
    'yes'
else
    'no'
end

But when I run the updated script, MATLAB displays an error message in the Command Window:

??? Operands to the || and && operators must be convertible to logical scalar values.

What can I do to fix this? Thanks in advance.

Andrew DeYoung

Carnegie Mellon University

[Sean de Wolski]

Your question is well written, you provide: an example, what you've tried, and what you want - everything we typically request. Good job.

[Andrew Newell]

I second that!

[Matt Fig]

This warning from MATLAB is silly. IF short circuits no matter whether you use & or &&. For example,

if false & error('Ooops')

disp('Here')

end

The only difference left is that & will accept vectors and && wants scalars. Note also that when using vectors, all elements must evaluate to true to pass the conditional.

You could also put this on the same line as the IF: %#ok

That will make M-Lint be quiet...

[Matt Tearle]

Discovered lurking in the doc for IF & "programming fundamentals -> basic program components -> operators":

"When used in the context of an if or while expression, and only in this context, the element-wise | and & operators use short-circuiting in evaluating their expressions. That is, A|B and A&B ignore the second operand, B, if the first operand, A, is sufficient to determine the result.

So, although the statement 1|[] evaluates to false, the same statement evaluates to true when used in either an if or while expression"

The idea of the warning (as I understand the reasoning) is that relying on conversion from array to scalar in an if/while is a bad idea. Eg "if x~=y" probably doesn't do what you'd expect when x and y are vectors. So the warning is essentially saying "hey, use && to be sure you're getting the behavior you want/expect". If you use &, your code might run, but not give intended behavior.

[Matt Fig]

But of course the same thing could be said for using any arrays with IF. All the time Answers we see folks puzzled by the output of their code when using vectors with IF. That all-or-none behavior (which essentially is scalar) with arrays gets a lot of people. Chaining this behavior with & makes little difference in this regard....

So I still say the warning is silly, unless M-Lint also gives a warning about IF with vector args (it doesn't).

[Matt Tearle]

You're right about IF and vectors, but the Code Analyzer doesn't necessarily know which variables are vectors and which aren't. The & operator is one instance where it can give a message without having to determine that. The str2num vs str2double message is another example. It's vaguely annoying to get a warning, but the Analyzer's just hedging its bets.

[BTW, I'm just passing on the "official" answer here. I was about to submit a request saying, basically, exactly what you've said. But I found an existing discussion, and I've paraphrased the end decision. I'm sure there's more to it than I can skim and pass along.]

[Jenny]

Matt, I still use my notes from your class, and thanks for the help in this question. I needed to get this right.

Answer 1

[Andrew Newell]

The problem is that size returns a vector:

size([1 2 3])

ans =

     1     3

Instead, use numel:

if numel([1 2 3])==numel([4 5 6]) && numel([4 5 6])==numel([7 8 9])
  disp('yes')
else
  disp('no')
end

Or you could use all(size([1 2 3])==size([4 5 6]) etc.

I have also put in the disp commands to take care of the other warnings.

[Andrew]

Thanks!

Answer 2

[Sean de Wolski]

As an addendum to Andrew's thorough and good solution, you can check the sizes directly:

if(isequal(size([1 2 3]),size([4 5 6])))
 disp('yes')
else
 disp('no')
end

This will fail if the sizes are not the same but the number of elements (numel) is:

if(isequal(size([1 2 3]),size([4; 5; 6])))
 disp('yes')
else
 disp('no')
end

[Andrew]

Thank you!

[Matt Tearle]

Another benefit to isequal is that it won't throw an error message in situations where == will (it will just return false). For example

if size(rand(2))==size(rand(3,2,2))

disp('yes')

else

disp('no')

end

will fail because size will return a 3-element vector for a 3-D array. However

if isequal(size(rand(2)),size(rand(3,2,2)))

disp('yes')

else

disp('no')

end

works fine.

[Andrew Newell]

Good point.