Confusion about elseif statements

I would like to generate code that decerns between input types scalar, vector (row or column), or matrix. I have written the following code:

function output = findtype(input);

[r c] = size(input);
if r == 1 && c == 1
    output = 'scalar';
elseif r == 1 || c == 1
    if r == 1
        output = 'row vector';
    else
        output = 'column vector';
    end
else
    output = 'matrix';
end
end

When I enter findtype(3), the code correctly returns 'scalar'. Likewise, when I enter findtype(ones(5,3)), the code correctly returns 'matrix'. However, when I enter any type of vector, whether it is findtype(1:5), findtype(5:1), or findtype (2:3), the code returns 'row vector'. It seems to me that in the second instance, the elseif statement should be true (since c = 1) and the if statement that follows should be false, since r ≠ 1, so the code should return 'column vector'. In the final instance, it seems the code should return 'matrix', since neither r nor c are equal to 1. I am confused as to why I am getting these results for vector inputs. Can someone help me understand where I've made an error? Thank you!

Answer 1

[Walter Roberson]

c will not be 1 for size([1:5]) . When you use the colon operator, the result is always a row vector, never a column vector.

Remember you are passing in the data whose size is to be checked, rather than passing in size information.

By the way, I recommend that you think more about how you want to handle empty values.

[Tim Robinson]

Yes, this was a very basic misunderstaninf on my part. Thank you for pointing this out. I checked my function with findtype(rand(10, 1)) and it works.

Answer 2

[KSSV]

The function is fine...there is one typo error..... you should use '=' not '=='

 function output = findtype(input);
[r c] = size(input);
if r == 1 && c == 1
    output = 'scalar';
elseif r == 1 || c == 1
    if r == 1
        output = 'row vector';
    else
        output = 'column vector';
    end
else
    output = 'matrix';
end
end

[Walter Roberson]

The function is not completely fine. Consider

findtype(rand(4,2,2))
findtype(rand(4,2,0))

[KSSV]

For that case, he can modify the line:

[r c] = size(input);

to

[r, c,d] = size(input);

He can follow the same check with r, c and make a new check with third dimension if he wants. As OP not mentioned about 3D arrays....I have not thought about that.

[Tim Robinson]

Thank you! I used = instread of == in my orginal function, but in my haste, I mistyped the function. I had not considered 3D arrays, but thank you for the suggestion!