why can't I get a vector as an output while using a constant in the function handle ?

I define the following constant function

f=@(x,y)0;

and I define the following vector

z=[1 2; 3 4; 5 6];

I want to evaluate the function f using feval

feval(f,z(:,1),z(:,2));

my question is that why can't I get the output as a vector instead of a scalar ?

Answer 1

[Azzi Abdelmalek]

Because that's what your function do, the result is always 0, it's independent of your inputs

[ahmad]

yes, but in the above example, i'm evaluating the function three times, so, I'm assuming i should get [0;0;0]

[Azzi Abdelmalek]

f=@(x,y)0; % the result is 0, doesn't depend on x and y.

you can use

f=@(x,y) zeros(size(x))

[Walter Roberson]

You are not evaluating the function three times: you are evaluating the function once, with a pair of vector arguments.

[ahmad]

but shouldn't I get the output in the same size as the input ?

[Azzi Abdelmalek]

I will give you a mathematical example

f(t)=10

The output is always 10, give any value for t, the output will not change

f(0)=10
f(100)=10

The same for your function

f=@(x,y)0

The result is independent of x and y, the result will be always 0. you can connect the size of your output to your input by

f=@(x,y) zeros(size(x))

or

f=@(x,y) x*0

Answer 2

[Walter Roberson]

f = @(x,y) zeros(size(x));

[Walter Roberson]

If you want to execute a function once for each input in a vector, you can use arrayfun

arrayfun(f, z(:,1), z(:,2))