Using switch to identify even or dd

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.

[MJFcoNaN]

Hello,

I will suggest you learn the basic programma of Matlab firstly. There may be significant discrepancy between languages.

[cgo]

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.

Answer 1

[Walter Roberson]

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

[cgo]

i am unaware of the 'true' statement there. why does it make the code work? Thanks for your insights.

[Walter Roberson]

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.

Answer 2

[KSSV]

@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

[cgo]

sorry, the problem insists on using switch. I am practising the command since I know if-else already.

[KSSV]

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

Answer 3

[David Hill]

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

[cgo]

sorry, the problem insists on using switch. I am practising the command since I know if-else already.