if-statement with conditions.

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KETAN PATEL
KETAN PATEL 2019 年 6 月 14 日
編集済み: DGM 2023 年 3 月 4 日
Write a function called picker that takes three input arguments called condition, in1 and in2 in this order. The argument condition is a logical. If it is true, the function assigns the value of in1 to the output argument out, otherwise, it assigns the value of in2 to out.
This is a problem from a course that I am taking on Coursera. I get a correct output for "true" but the output for the "false" condition is always wrong. Please, any kind of help will highly appreciated.
Annotation 2019-06-14 203145.png
Here is my code:
function out = picker(condition,in1,in2)
if true
out = in1;
else
out = in2;
end
end
  8 件のコメント
Harish C S
Harish C S 2021 年 3 月 29 日
this is wrong because everytime matlab thinks it is true
Rik
Rik 2021 年 3 月 29 日
@Harish C S That is why it is a question. If it were correct, why bother posting the question? See the comments and answers in this thread for working code.

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採用された回答

Star Strider
Star Strider 2019 年 6 月 14 日
You can safely delete this assignment:
condintion = in1 < in2;
because it overwrites the ‘condition’ argument.
You can also just use:
if conditon
You don’t have to test it again.
Also, check your spelling!
  4 件のコメント
Priya Narayan
Priya Narayan 2020 年 6 月 17 日
This code is not running.
Rik
Rik 2020 年 6 月 17 日
Which code? And as Star Strider suggests: check your spelling, including in the code he posted.

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その他の回答 (4 件)

Sneham Shrikant Vaidya
Sneham Shrikant Vaidya 2020 年 7 月 18 日
function out = picker(condition,in1,in2)
if condition == 1;
out = in1;
else
out = in2;
end
\\try this above code

Ahmed Salmi
Ahmed Salmi 2020 年 7 月 17 日
function out=picker(condition,in1,in2)
if condition==true
out = in1;
elseif condition==false
out = in2;
end
end
  1 件のコメント
Walter Roberson
Walter Roberson 2020 年 8 月 31 日
Under what circumstances can it be the case that the condition is not true and also is not false? When you use elseif you imply that there are cases where nothing that has been tested before came out true and also that the current test might not come out true either.

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suat karabocek
suat karabocek 2019 年 6 月 20 日
you may use if and else and two conditions including 1 and 0. such as;
your function.........
if condition == 1;
....
....
else condition == 0;
....
....
end
  4 件のコメント
Ahmed J. Abougarair
Ahmed J. Abougarair 2020 年 4 月 19 日
The picker function required three input argument
DGM
DGM 2023 年 3 月 4 日
編集済み: DGM 2023 年 3 月 4 日
You may use two if conditions with numeric comparison, but why should you?
if condition == 1
out = in1;
elseif condition == 0
out = in2;
% else
% all other cases are unhandled and will cause the function
% to exit without assigning a value to the output
end
The assignment asserts that condition is a variable of class 'logical'. While the comparisons using numeric values will implicitly work with logical inputs, you're designing your code around handling logical variables as numeric variables. Your code now accepts any numeric input and will fail if it is neither 0 nor 1.
If your inputs are supposed to be logical, make sure they're logical or converted to logical.
Also, an else statement doesn't accept a condition; an elseif does.

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amjad khan
amjad khan 2020 年 4 月 9 日
編集済み: Walter Roberson 2020 年 8 月 31 日
function out(condition,in1,in2)
if condition>0 % it is logical
out=in1;
else
out=2;
end
  1 件のコメント
DGM
DGM 2023 年 3 月 4 日
The only significant difference between this code and the other examples is that this one doesn't actually work. It may be a minor and obvious error, but the fact that it's obvious indicates that this wasn't tested.
Why post untested code that's not meaningfully different than other examples?
You do have one comment in the code, so I'll give you credit for that.

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