# Using AND Operator in “if” statements

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Andrew on 10 May 2011
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
##### 7 CommentsShow 5 older commentsHide 5 older comments
Matt Tearle on 11 May 2011
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 on 2 Aug 2016
Matt, I still use my notes from your class, and thanks for the help in this question. I needed to get this right.

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### Accepted Answer

Andrew Newell on 10 May 2011
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.
##### 1 CommentShow -1 older commentsHide -1 older comments
Andrew on 10 May 2011
Thanks!

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### More Answers (1)

Sean de Wolski on 10 May 2011
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
##### 3 CommentsShow 1 older commentHide 1 older comment
Matt Tearle on 10 May 2011
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 on 10 May 2011
Good point.

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