A problem with function handle
Ältere Kommentare anzeigen
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
Akzeptierte Antwort
Weitere Antworten (0)
Kategorien
Mehr zu Data Type Identification finden Sie in Hilfe-Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!