Using switch to identify even or dd
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cgo
am 20 Jul. 2022
Kommentiert: Walter Roberson
am 20 Jul. 2022
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 Kommentare
MJFcoNaN
am 20 Jul. 2022
Hello,
I will suggest you learn the basic programma of Matlab firstly. There may be significant discrepancy between languages.
Akzeptierte Antwort
Walter Roberson
am 20 Jul. 2022
Bearbeitet: Walter Roberson
am 20 Jul. 2022
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 Kommentare
Walter Roberson
am 20 Jul. 2022
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.
Weitere Antworten (2)
KSSV
am 20 Jul. 2022
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 Kommentare
KSSV
am 20 Jul. 2022
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
David Hill
am 20 Jul. 2022
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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