Re: if statement within function

[Brian Blais]

Ahn Kyung wrote:
> I'm having trouble implementing if statement in a function. Here's an 
> example:
> ---------------------
> function y=f(x)
>  if (x>1)
>    y=1;
>  else
>    y=x.^2;
>  endif
> endfunction
> 
> x=[1;2];
> 
> 
> octave:3> f(x)
> ans =
> 
>  1
>  4
> 
> octave:4> f(1)
> ans = 1
> octave:5> f(2)
> ans = 1
> --------------------
> You see? f(2)=1, as expected, but if I use a vector x=[1;2];, I get 
> f(x)=[1;4]; When using a vector, the if statement is useless...
> 
> I don't want to pass each element, but want to use f(x) as a chunk, with 
> a given vector x. What should I do to cure this problem??
> 

If I am understanding the issue here, you wanted f([1;2]) to return
[1;1] not [1;4], right?

If that is true, then you are misunderstanding the if-statement.  It
checks only 1 true/false value, and doesn't change the program flow for
each element of the matrix.  If you want the effect you state, you first
need to find all of the elements of your vector x which are greater than
1, and those less than 1, and deal with it that way.  For example,

 function y=f(x)

y=zeros(size(x));  % make a return vector the same size as x

idx=find(x>1);
y(idx)=1;   % all those elements greater than 1 are set to 1

idx=find(x<=1);
y(idx)=x(idx).^2;   % all those elements less than one

endfunction

now I get:

>> x=[1;2];
>> f(x)
ans =

  1
  1



it's a different way of thinking about the program flow, when you are
dealing with vectors.

I hope this helps,

                        Brian Blais

-- 
-----------------

            address@hidden
            http://web.bryant.edu/~bblais



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