A problem with function handle
Mostrar comentarios más antiguos
Hello,
I have a simple problem but I do not understand why function handle behaves like this!!! I explain by a simple example in bellow:
>> x1=linspace(-pi,pi,20);y1=sin(x1);f=@(x)interp1(x1,y1,x);f(5)
ans =
NaN
which makes sense. But, when I type
>> x1=linspace(-pi,pi,20);y1=sin(x1);f=@(x)interp1(x1,y1,x).*(x>=-pi & x<=pi)+0.*(x<-pi | x>pi);f(5)
ans =
NaN
and this does not make sense to me.
Any idea?
Thanks in advance,
Babak
Respuesta aceptada
Más respuestas (0)
Categorías
Más información sobre Data Type Identification en Centro de ayuda y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!