Using switch to identify even or dd

137 次查看（过去 30 天）
cgo
cgo 2022-7-20
评论： Walter Roberson 2022-7-20
Problem in brief: Print “Odd” if the argument is 1, 3, or 5, “Even” if the argument is 0, 2, or 4, and “Let me get back to you on that one.” for any other value.
function [y] = even_odd(x)
r = mod(x,2);
switch x
case x<=4 && r ==0
fprintf('even \n')
case x<=5 && r == 1
fprintf('odd \n')
otherwise
fprintf('i will get back to you on that \n')
end
end
I dont understand why this function doesn't give the correct response. Please provide insights.
  2 个评论
MJFcoNaN
MJFcoNaN 2022-7-20
Hello,
I will suggest you learn the basic programma of Matlab firstly. There may be significant discrepancy between languages.
cgo
cgo 2022-7-20
this is matlab language. I want to know if there is a mistake in the logic as I am unable to get the intended outputs.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2022-7-20
编辑：Walter Roberson 2022-7-20
Ran in:
even_odd(-88)
even
even_odd(2)
even
even_odd(3)
odd
even_odd(7)
i will get back to you on that
function [y] = even_odd(x)
r = mod(x,2);
switch true
case x<=4 && r ==0
fprintf('even \n')
case x<=5 && r == 1
fprintf('odd \n')
otherwise
fprintf('i will get back to you on that \n')
end
end
  2 个评论
cgo
cgo 2022-7-20
i am unaware of the 'true' statement there. why does it make the code work? Thanks for your insights.
Walter Roberson
Walter Roberson 2022-7-20
Ran in:
The value that you list in the switch statement is compared to the value listed in the case. The values in your case are things like x<=4 && r==0 which is a logical expression, so the values in your case are either true or false so you need to switch on one of true, false, 0, or 1 . You want to select the case that is true, so you have to switch on true
... in practice you do not need that comparison for x inside the switch. You might hypothetically want to test against
fprintf('%ld\n', flintmax)
9007199254740992
which is the largest double precision integer that can be reliably tested for division by 2.

请先登录，再进行评论。

更多回答（2 个）

KSSV
KSSV 2022-7-20
@cgo
You have only two options x can be either even or odd..that's all.
x = 4 ;
r = mod(x,2);
if r == 0
fprintf('%d is Even\n',x)
else
fprintf('%d is Odd\n',x)
end
  2 个评论
cgo
cgo 2022-7-20
sorry, the problem insists on using switch. I am practising the command since I know if-else already.
KSSV
KSSV 2022-7-20
With switch, case, otherwise
x = 5 ;
r = mod(x,2);
switch r
case 1
fprintf('%d is Odd\n',x)
case 0
fprintf('%d is Even\n',x)
end

请先登录，再进行评论。

David Hill
David Hill 2022-7-20
Look at the function switch, case, otherwise, it does not work as you coded.
function [y] = even_odd(x)
r = mod(x,2);
if x<=4 && r==0
fprintf('even \n')
elseif x<=5 && r==1
fprintf('odd \n')
else
fprintf('i will get back to you on that \n')
end
end
  1 个评论
cgo
cgo 2022-7-20
sorry, the problem insists on using switch. I am practising the command since I know if-else already.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Data Type Conversion 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by