if-statement with conditions.

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 个评论

This code gives error
Assessment result: incorrectFalse
Variable out has an incorrect value.
function out = picker(condition,in1,in2)
if condition == 1;
out = in1;
else
out = in2;
end
\\try this code
function out=picker(condition,in1,in2)
if condition==true
out=in1;
else
out=in2;
end
it is a simplest code that can execute your program.
No, that is not the simplest ;-)
Jaideep has written the right code
this is wrong because everytime matlab thinks it is true
@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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 采纳的回答

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 个评论

Correct code is:
function out = picker(condition,in1,in2)
if condition
out = in1;
else
out = in2;
end
end
This code is not running.
Which code? And as Star Strider suggests: check your spelling, including in the code he posted.

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更多回答(4 个)

function out = picker(condition,in1,in2)
if condition == 1;
out = in1;
else
out = in2;
end
\\try this above code
function out=picker(condition,in1,in2)
if condition==true
out = in1;
elseif condition==false
out = in2;
end
end

1 个评论

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 2019-6-20

0 个投票

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 个评论

function out = picker(a,b)
if(a<b)
fprintf("%d",a);
else
fprintf("%d",b);
end
what the problem on this code
The picker function required three input argument
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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function out(condition,in1,in2)
if condition>0 % it is logical
out=in1;
else
out=2;
end

1 个评论

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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