How to get rid of Warning"Unexpected use of '[' in a scalar context."

42 vues (au cours des 30 derniers jours)
Crystal Lee
Crystal Lee le 13 Août 2018
Réponse apportée : David le 7 Déc 2024
a part of the function has:
if the input A (a vector) matches what is stated, the statement runs.
if A == [1,3,3]
% statement
end
and it gives me this error: 'Unexpected use of '[' in a scalar context. How can I fix this?
  2 commentaires
dpb
dpb le 13 Août 2018
Must be more than that; that code runs w/o any error here...
>> A=3;
>> if A==[1, 3, 3],disp('ok'),end
>> A=[1 3 3];
>> if A==[1, 3, 3],disp('ok'),end
ok
>>
Show us the context including the error message with all the text generated...
Crystal Lee
Crystal Lee le 13 Août 2018
There is no red text, I was just wondering if I could get rid of the orange warning.

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Réponse acceptée

Stephen23
Stephen23 le 13 Août 2018
Modifié(e) : Stephen23 le 13 Août 2018
Use all or any, like this:
if all(A==[1,3,3])
Or right-click the pop-up warning box and follow the instructions to suppress it.
  3 commentaires
Thiago  Rodrigues
Thiago Rodrigues le 17 Sep 2020
What the explanation for the "all" fix the warning?
Stephen23
Stephen23 le 17 Sep 2020
Modifié(e) : Stephen23 le 17 Sep 2020
"What the explanation for the "all" fix the warning?"
Given a vector input the output from all is scalar. A scalar condition is easier to understand, which is why it is recommended by that warning.

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Plus de réponses (2)

KSSV
KSSV le 13 Août 2018
When you use A == [1 3 3] ;
With this operator ==, MATLAB expects a scalar value, but the way used here, it gives array as output. So the warning pops out. The way you use, won't work out. YOu need to use the operator == with a scalar value, so the if condition works fine.
  5 commentaires
KSSV
KSSV le 13 Août 2018
Yes....it is not a error..it is a warning....I am using 2015b.
Crystal Lee
Crystal Lee le 13 Août 2018
Sorry, I confused warning with error.

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David
David le 7 Déc 2024
You may want to use isequal instead of == for that use case.

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