Effacer les filtres
Effacer les filtres

if-statement with conditions.

16 vues (au cours des 30 derniers jours)
KETAN PATEL
KETAN PATEL le 14 Juin 2019
Modifié(e) : DGM le 4 Mar 2023
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 commentaires
Harish C S
Harish C S le 29 Mar 2021
this is wrong because everytime matlab thinks it is true
Rik
Rik le 29 Mar 2021
@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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Réponse acceptée

Star Strider
Star Strider le 14 Juin 2019
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 commentaires
Priya Narayan
Priya Narayan le 17 Juin 2020
This code is not running.
Rik
Rik le 17 Juin 2020
Which code? And as Star Strider suggests: check your spelling, including in the code he posted.

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Plus de réponses (4)

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

Ahmed Salmi
Ahmed Salmi le 17 Juil 2020
function out=picker(condition,in1,in2)
if condition==true
out = in1;
elseif condition==false
out = in2;
end
end
  1 commentaire
Walter Roberson
Walter Roberson le 31 Août 2020
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 le 20 Juin 2019
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 commentaires
Ahmed J. Abougarair
Ahmed J. Abougarair le 19 Avr 2020
The picker function required three input argument
DGM
DGM le 4 Mar 2023
Modifié(e) : DGM le 4 Mar 2023
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 le 9 Avr 2020
Modifié(e) : Walter Roberson le 31 Août 2020
function out(condition,in1,in2)
if condition>0 % it is logical
out=in1;
else
out=2;
end
  1 commentaire
DGM
DGM le 4 Mar 2023
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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