does anyone know how to rewrite this into switch/case statements

n=input('Please enter a positive or negative integer: ');
if n < -2 || n > 5
    disp(n/2)
else
    if n <= 3
        if n >= 0
            disp(floor(n/2))
        else
            disp(ceil(n))
        end
    else
        disp(1)
    end
end

[Rik]

Why do you want to put this into a switch/case? It doesn't look really suited to do so.

[John D'Errico]

Please don't ask the same question multiple times. It will only get the others closed anyway.

Regardless, Rik is completely correct here. Trying to write this as a switch/case is trying to fit a square peg into a round hole. You can probably do it with a large enough hammer, but why?

You have code that works, does what you want to do.

[Pablo Garcia]

its a hw question for my engineering class. Im dying trying to figure this out

[Pablo Garcia]

sorry for the mistake John, I made a mistake on my first question so I posted another. I also forgot to comment how I am not allowed to use the if or if-else statement

[John D'Errico]

Yeah, I figured it was HW. Stupid stuff. Teacher makes you do silly things for no good reason. Grump. Rant. Grumble. ;-)

So, are you asking how to do different things, based on which range n falls in, using a switch/case?

[Pablo Garcia]

The homework is basically just asking us to switch the above code from an if/if-else statement to a switch/case statement that still keeps the same purpose. I understand how to use switch/case statements but I'm having trouble with the inequalities

Answer 1

[John D'Errico]

The immediate answer is to look at histcounts. It will allow you to set up a series of break points, returning an integer that indicates which regime a number falls into. So it looks like the various cases you might care about are:

breaks = [-2 0 3 5]

Then the third output from histcounts will be an integer, depending on which interval a point falls in. The problem is, you have a special issue, in that histcounts tests to see if a point is in an interval if it is >= the breaks.

So histc would put 3 and 4 into the same bin! And it looks like you wants histcounts to do something different for n==3 versus n==4.

If you know that n is always an integer, that means you wanted to set the break points as

breaks = [-2 0 4 5];

Anyway, histcounts may be beyond the scope of this problem, since it has a simple resolution.

switch n
  case {-1 -2}
    % do whatever you might do when n==-1 or -2

    case {0 1 2 3}
      % do whatever you might do when n lives in this range

    case {4 5}
      % do whatever you might do when n lives in this range

    otherwise
      % we covered all the other cases, so this is what happens when n < -2 or > 5
      disp(n/2)

end

[Pablo Garcia]

How would I write a case for n>5 and n<-2, which values would be divided by n/2 in that range?

[Bruno Luong]

"otherwise"

[John D'Errico]

That is exactly what otherwise does. It covers all cases that were not explicitly mentioned in the cases. The problem would be if you wanted to do something different for n<-2 compared to n>5. Then you would need to get trickier. For example, you might need to put an if inside the otherwise block.