Using logical operator in an if condition

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Kaan Yilmaz
Kaan Yilmaz 2016 年 6 月 5 日
回答済み: aakash jain 2021 年 3 月 4 日
Hi, I want my 'if' condition as follows:
if true
if (x==1) && (A(2,1)==1 || A(2,2)==1 || A(2,3)==1)
...
end
(Please disregard the first line with 'if true' )
I mean, I want my condition to be x equals 1 and one of A(2,1) A(2,2) or A(2,3) to be equal 1 as well. I think, the way that the expression above is written, is incorrect because even the condition isn't satisfied, program continues to execute the if statement.
How should I write the above condition?
Thank you Kaan Yilmaz

回答 (2 件)

Roger Stafford
Roger Stafford 2016 年 6 月 5 日
編集済み: Roger Stafford 2016 年 6 月 5 日
You are using the "short circuit" logical operators. They work for scalars but not for vectors with more than one element. In the latter case change to '&' and '|'.
Another possibility for your difficult is that you are requiring exact equality which may not be satisfied if x, y, z, or t differ from 1 in their least binary digits from rounding errors.
  2 件のコメント
Kaan Yilmaz
Kaan Yilmaz 2016 年 6 月 5 日
編集済み: Kaan Yilmaz 2016 年 6 月 5 日
I wrote y z t for simplicity but they are actually elements of a matrix and they are like A(2,1) A(2,2) and A(2,3). Their values are not computed but assigned by a function so a rounding error can not be a consideration.
What is unexpected here is that even all of the elements of the A matrix are 0, the code pretends like the condition is satisfied which requires at least some elements to be 1.
I also tried to use & and | but it was not a solution at all.
Thank you,
Roger Stafford
Roger Stafford 2016 年 6 月 5 日
編集済み: Roger Stafford 2016 年 6 月 5 日
I would like to see your actual code that you are using to produce the wrong result, including how A is generated.

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aakash jain
aakash jain 2021 年 3 月 4 日
Error in untitled1233 (line 4)
if Z == 0 || X==P
ERROR

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