# Using switch to identify even or dd

조회 수: 1(최근 30일)
cgo 2022년 7월 20일
댓글: Walter Roberson 2022년 7월 20일
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.
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
cgo 2022년 7월 20일
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.

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### 채택된 답변

Walter Roberson 2022년 7월 20일
편집: Walter Roberson 2022년 7월 20일
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
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
Walter Roberson 2022년 7월 20일
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.

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### 추가 답변(2개)

KSSV 2022년 7월 20일
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
##### 댓글 수: 2표시숨기기 이전 댓글 수: 1
KSSV 2022년 7월 20일
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

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David Hill 2022년 7월 20일
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
##### 댓글 수: 1표시숨기기 없음
cgo 2022년 7월 20일
sorry, the problem insists on using switch. I am practising the command since I know if-else already.

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