How can I tell if this is stored as a vector?

Basically, I created a function, which contains the following (along with some other stuff):

Lala= myfunction(a,b,n)
...

LaLa=zeros(1,n);
index= 3:n
Lala(index)= Lala(index-1) + Lala(index-2);

How can I tell if I'm just "displaying the series elements as I calculate them" or if I'm "storing and displaying the series as a vector"?

This is a homework assignment where they asked us to first display the first n numbers of the series, and then it says to 'modify the function' to return the series as a vector.

I've tried doing this: Lala= [1:length(n)]; (instead of the zeros function).

However, I don't know how to tell if Lala is a vector or not? It displays the same either way: 0 1 1 2

Answer 1

[Stephen23]

"Displaying" in this context typically means to print some data to the command window. There are several ways to achieve this, such as:

• any expression will display its output when it does NOT have a trailing semicolon.
• calling disp with some input data
• calling fprintf with some input data
• error or warning calls.

So to fulfill the first part of the assignment "...first display.." you would need something like this:

function myfunction(a,b,n) % no output argument!
...
LaLa  = zeros(1,n);
index = 3:n;
Lala(index)= Lala(index-1) + Lala(index-2) % no semicolon!

and for the second part you can add the semicolon (so it does NOT display any more) and add the output argument so that it returns those values:

function LaLa = myfunction(a,b,n) % with output argument!
...
LaLa  = zeros(1,n);
index = 3:n;
Lala(index)= Lala(index-1) + Lala(index-2); %semicolon!

"However, I don't know how to tell if Lala is a vector or not?"

A vector has size 1xN or Nx1. According to that LaLa is a vector, because that is how you defined it using zeros(1,n). If n is a scalar then 1:length(n) is also a scalar (you did not tell us what size n is).

Answer 2

[Rachel Dawn]

Wow, thank you so much! I wouldn't have thought it would be as simple as taking off a semicolon & the output argument. I thought it had to be something more complicated than that that I was missing.

You're a life saver!

[Stephen23]

@Rachel Dawn: I hope that it helped. You should also accept the answer that best helped resolve your original question: this tells other users that your question has been resolved.