Using switch to identify even or dd
显示 更早的评论
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.
采纳的回答
even_odd(-88)
even_odd(2)
even_odd(3)
even_odd(7)
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
2022-7-20
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)
which is the largest double precision integer that can be reliably tested for division by 2.
更多回答（2 个）
KSSV
2022-7-20
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
David Hill
2022-7-20
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
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